In the previous section, we saw 'completely inelastic collision'.
• The collisions that we saw in the previous section and in the section before that, were all one-dimensional collisions. They are also called head-on collisions
• For a collision to be classified as a one-dimensional collision, some conditions are to be satisfied. In this section, we will see those details
1. Consider two bodies A and B in fig.6.69 below. Their initial and final velocities are shown in the fig.
• Before collision, the velocities are: vAi and vBi
• After collision, the velocities are: vAf and vBf
■ If all the four velocities lie along the same straight line, that collision is called a one-dimensional collision
2. Note that, the vectors indicating the velocities are all drawn through the centers of the bodies.
3. Also note the difference in levels:
♦ The 'top of A' is at a different level from 'top of B'
♦ The 'bottom of A' is at a different level from 'bottom of B'
• If any one of the above pairs of levels are the same, we will not get a one-dimensional collision
• So the two bodies cannot be moving on the top of a table. Because, if they do so, then their bottom levels will be the same, and the the collision will not be one-dimensional
• Such a problem arises because of the 'difference in the sizes' fo the two bodies
[In fig.6.69, we are not concerned about the directions (towards left or towards right) of the velocities. Those directions will depend on the magnitudes of initial velocities and the masses of the two bodies. What is important here is that, all the velocities are along the same line]
4. The simplest case which satisfies condition in (1) is shown in fig.6.70 below:
• The two spheres A and B have the same diameter
• They rest on a smooth horizontal surface
• Sphere B is initially at rest
■ If vAi passes through the center of sphere B, the collision that takes place will be a one-dimensional collision
Let us see two more solved examples:
Solved example 6.37
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v. If the collision is elastic, which of the following (fig. 6.71) is a possible result after collision?
Solution:
• This problem can be solved by two methods. Before using the easy method, we will see the normal method:
1. Before collision, we have two balls 'B' and 'C' in contact with each other
• Given that 'B' and 'C' are identical. So they have the same diameter
2. Ball 'A' collides elastically with 'B'
• After the collision between 'A' and 'B', the second ball 'B' will certainly move and collide with 'C'
• Given that the collisions are head-on. That is., the collisions are one dimensional
3. First let us consider the collision between 'A' and 'B'
• Since all balls are identical, we can write mA = mB = mC = m
• If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{mv+m \times 0=m v_{Af}+m v_{Bf}}$
• Substituting the values, we get:
$\mathbf\small{\Rightarrow v= v_{Af}+ v_{Bf}}$
4. In an elastic collision, kinetic energy is conserved. So we have:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Substituting the values, we get:
$\mathbf\small{mv^2+m \times 0^2=m v_{Af}^2+m v_{Bf}^2}$
$\mathbf\small{\Rightarrow v^2=v_{Af}^2+v_{Bf}^2}$
5. Squaring the result in (3), we get: $\mathbf\small{v^2= v_{Af}^2+2v_{Af}\;v_{Bf}+ v_{Bf}^2}$
Substituting for $\mathbf\small{v^2}$ from(4), we get: $\mathbf\small{v_{Af}^2+ v_{Bf}^2= v_{Af}^2+2v_{Af}\;v_{Bf}+ v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0= 2v_{Af}\;v_{Bf}}$
• Final velocity of B (vBf) cannot be zero because, it is hit by the ball 'A' of equal mass
• And also, there is no friction to stop 'B'
• So we get vAf = 0
6. Substituting this value of vAf in (3), we get: vBf = v
• So we can write a summary about the collision between A and B:
(i) Two balls 'A' and 'B' collide elastically with each other
(ii) The two masses are equal. That is., mA = mB = m
(iii) Before the collision, 'A' was moving with a velocity v
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
(v) After the collision, 'A' becomes stationary
(vi) After the collision, 'B' begins to move with v (the same velocity with which 'A' was moving)
7. Now we consider the collision between 'B' and 'C'
• B has an initial velocity of 'v' and C is stationary
• So this is similar to the collision between A and B
• The steps will be same as from (3) to (6) above
• We can write a summary about the collision between B and C:
(i) Two balls 'B' and 'C' collide elastically with each other
(ii) The two masses are equal. That is., mB = mC = m
(iii) Before the collision, 'B' was moving with a velocity v
(iv) Before the collision, 'C' was stationary. That is., vCi = 0
(v) After the collision, 'B' becomes stationary
(vi) After the collision, 'C' begins to move with v (the same velocity with which 'B' was moving)
8. So we see that, out of the 3 options given on the right side in fig.6.71 above, (ii) is the correct option
Easy method:
1. We have derived formulas for vAf and vBf in the case of elastic one-dimensional collision between two objects A and B (with B initially at rest)
• The formulas are:
$\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
$\mathbf\small{v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
(Details here)
2. Further, we saw special case 1 in which the two objects have the same mass
• The collision between balls A and B in this problem, will fall under this case 1
• So we can directly write the result:
After the collision, 'A' will come to a stop and 'B' will begin to move with velocity 'v'
3. The collision between 'B' and 'C' also falls under this case 1
• So we can directly write the result:
• After the collision, 'B' will come to a stop and 'C' will begin to move with velocity 'v'
4. So we see that, out of the 3 options given on the right side in fig.6.71 above, (ii) is the correct option
Solved example 6.38
The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in fig. 6.72(a). How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic
Solution:
1. The bob B is kept vertically below the point O
• The bob B is on the path of A. This is shown in fig.b
• So when the bob A hits B, it will be a one-dimensional collision
2. Given that, the two bobs have the same mass
3. Given that the collision is elastic
4. Given that, the size of the bobs can be neglected
• So we can assume that they have the same diameter
5. Given that B is at rest
6. Based on the five points above, the collision falls under special case 1
• We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• In special case 1, mA = mB
• This gives vAf = 0
• So we can directly write the result:
■ After the collision, 'A' will come to a stop and 'B' will begin to move with velocity 'v'
♦ Where 'v' is the velocity attained by the bob A when it reaches the lowest point
7. Thus we see that, the bob A will not rise to any height. It will come to a stop at the very instant when collision occurs
• We will get the same result even if we change the angle 30o to any other value
While performing a collision experiment on a horizontal surface, we face three difficulties. They are:
1. There will be friction between the colliding bodies and the surface
• So some energy will be lost for doing work against friction
• As a result, their will be retardation (negative acceleration) and so the bodies will not be traveling with uniform velocities
• So we cannot write the conservation of kinetic energy equation
2. If the colliding bodies have different sizes, the velocities will not fall along the same straight line. So we will not get a one-dimensional collision
• This is shown in fig.6.73 below:
3. Since the velocities vary, it is difficult to obtain the exact velocities just before (vi) and after collision (vf)
1. Take a large ball like a basketball/football/volleyball
• Let us take a basketball. It has a mass mB = 0.58 kg
2. Allow it to fall freely from a convenient height (h1) of say 1.5 m. This is shown in fig.6.74(a) below:
• Measure the height (h2) to which it rebounds. Let it be 1.2 m. This is shown in fig.b
3. We want the velocity (v1) with which it hits the ground
• That is., we want the velocity just before collision
• We can easily obtain it using the familiar equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
$\mathbf\small{v_1^2=2\times g \times h_1=2\times 9.8 \times 1.5=29.4}$
$\mathbf\small{\therefore v_1=\sqrt{29.4}=5.422\;\;\text{ms}^{-1} }$
4. We want the velocity (v2) with which it leaves the ground after rebound
• That is., we want the velocity just after collision
• We can easily obtain this also using the same equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
• $\mathbf\small{v_2^2=2\times g \times h_2=2\times 10 \times 1.2=23.52}$
• $\mathbf\small{\therefore v_2=\sqrt{23.52}=4.849\;\;\text{ms}^{-1} }$
5. Once we determine v1 and v2, we can obtain the coefficient of restitution using the relation:
$\mathbf\small{\text{Coefficient of restitution}\;\;(e)=\frac{\text{Velocity after collision}}{\text{Velocity before collision}}=\frac{v_2}{v_1}=\frac{4.849}{5.422}=0.894}$
6. Now take a smaller tennis ball/rubber ball/ table tennis ball
• Let us take a tennis ball. It has a mass mT = 0.057 kg
7. While holding the basketball, stack the tennis ball vertically above it
• Drop them together from the same height h1 of 1.5 m. This is shown in fig.6.74(c)
• Following two points must be ensured while making the drop:
(i) The two balls must move together
(ii) The smaller ball must be exactly vertical above the larger ball
• Then only we will get a one-dimensional collision
8. Measure the height (h3) upto which the basketball rebounds
Let h3 = 0.75 m. This is shown in fig.d
9. After collision with the ground, we find that:
• The tennis ball rebounds to a larger height (h4). This is shown in fig.e
• It is not easy to measure that h4
• So we will use analytical method to find h4
• While using the analytical method, we will see the applications of:
♦ Conservation of momentum
♦ Conservation of energy
10. Note that, the basketball does not rebound to the same height obtained earlier
That is., h3 < h2
11. Let us analyse what happens during the collision:
• There are actually two collisions
(i) The first one is that between the basketball and the floor
(ii) As a result of that collision, the basket ball rebounds and collides with the tennis ball which was trailing close behind. This is the second collision
■ This second collision is the one which is of interest to us
12. For this second collision, we have:
• mB = 0.58 kg, mT = 0.057 kg
• vBi = v2 = 4.849 ms-1
[∵ the basket ball will have a rebound velocity of 4.849 ms-1 after hitting the floor. This we obtained in (4)]
• vTi = v1 = -5.422 ms-1
[∵ any mass falling through 1.5 m, will have a velocity of 5.422 ms-1. This we obtained in (3)
The negative sign is given because it is in the downward direction]
• vBf = ?, vTf = ?
13. Let us apply the principle of conservation of momentum:
$\mathbf\small{m_B\,v_{Bi}+m_T\,v_{Ti}=m_B\,v_{Bf}+m_T\,v_{Tf}}$
• Substituting the known values, we get:
$\mathbf\small{0.58\times 4.849-0.057\times 5.422=0.58\,v_{Bf}+0.057\,v_{Tf}}$
$\mathbf\small{\Rightarrow 0.58\,v_{Bf}+0.057\,v_{Tf}=2.503}$
$\mathbf\small{\Rightarrow 10.175\,v_{Bf}+\,v_{Tf}=43.926}$
$\mathbf\small{\Rightarrow v_{Tf}=43.926-10.175\,v_{Bf}}$
14. Let us apply the principle of conservation of kinetic energy:
• Assuming an elastic collision, we have:
$\mathbf\small{\frac{1}{2}m_B\,v_{Bi}^2+\frac{1}{2}m_T\,(-v_{Ti})^2=\frac{1}{2}m_B\,v_{Bf}^2+\frac{1}{2}m_T\,v_{Tf}^2}$
• Multiplying both sides by '2', we get:
$\mathbf\small{m_B\,v_{Bi}^2+m_T\,v_{Ti}^2=m_B\,v_{Bf}^2+m_T\,v_{Tf}^2}$
• Substituting the known values, we get:
$\mathbf\small{13.642+1.676=0.58\,v_{Bf}^2+0.057\,v_{Tf}^2}$
$\mathbf\small{\Rightarrow 15.317=0.58\,v_{Bf}^2+0.057\,v_{Tf}^2}$
15. So we have two equations:
(i) From (13) we have: $\mathbf\small{v_{Tf}=43.926-10.175\,v_{Bf}}$
(ii) From (14) we have: $\mathbf\small{0.58\,v_{Bf}^2+0.057\,v_{Tf}^2=15.317}$
• Solving them, we get: vBf = 4.853 OR 3.01
• When vBf = 4.853, we get vTf = -5.453 ms-1
• When vBf = 3.01, we get vBf = 13.3 ms-1
16. The negative velocity is not acceptable because, the tennis ball moves upwards after collision
■ So we can write:
After the collision, the tennis ball moves upwards with the velocity vTf = 13.3 ms-1
17. The height upto which the tennis ball reaches can be calculated using the familiar equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
$\mathbf\small{v_{Tf}^2=2\times g \times h_4}$
$\mathbf\small{\Rightarrow 13.3^2=2\times 9.8 \times h_4}$
$\mathbf\small{\Rightarrow h_4=9.025\;\text{m}}$
■ 13.3 ms-1 is a very high velocity. And 9.025 m is nearly 3 storeys high. So all necessary safety precautions should be adopted while performing this experiment
• The collisions that we saw in the previous section and in the section before that, were all one-dimensional collisions. They are also called head-on collisions
• For a collision to be classified as a one-dimensional collision, some conditions are to be satisfied. In this section, we will see those details
1. Consider two bodies A and B in fig.6.69 below. Their initial and final velocities are shown in the fig.
 |
| Fig.6.69 |
• After collision, the velocities are: vAf and vBf
■ If all the four velocities lie along the same straight line, that collision is called a one-dimensional collision
2. Note that, the vectors indicating the velocities are all drawn through the centers of the bodies.
3. Also note the difference in levels:
♦ The 'top of A' is at a different level from 'top of B'
♦ The 'bottom of A' is at a different level from 'bottom of B'
• If any one of the above pairs of levels are the same, we will not get a one-dimensional collision
• So the two bodies cannot be moving on the top of a table. Because, if they do so, then their bottom levels will be the same, and the the collision will not be one-dimensional
• Such a problem arises because of the 'difference in the sizes' fo the two bodies
[In fig.6.69, we are not concerned about the directions (towards left or towards right) of the velocities. Those directions will depend on the magnitudes of initial velocities and the masses of the two bodies. What is important here is that, all the velocities are along the same line]
4. The simplest case which satisfies condition in (1) is shown in fig.6.70 below:
 |
| Fig.6.70 |
• They rest on a smooth horizontal surface
• Sphere B is initially at rest
■ If vAi passes through the center of sphere B, the collision that takes place will be a one-dimensional collision
The solved examples that we saw in the previous two sections were all one-dimensional collisions.
Solved example 6.37
Two identical ball bearings in contact with each other and resting on a frictionless table are hit head-on by another ball bearing of the same mass moving initially with a speed v. If the collision is elastic, which of the following (fig. 6.71) is a possible result after collision?
 |
| Fig.6.71 |
• This problem can be solved by two methods. Before using the easy method, we will see the normal method:
1. Before collision, we have two balls 'B' and 'C' in contact with each other
• Given that 'B' and 'C' are identical. So they have the same diameter
2. Ball 'A' collides elastically with 'B'
• After the collision between 'A' and 'B', the second ball 'B' will certainly move and collide with 'C'
• Given that the collisions are head-on. That is., the collisions are one dimensional
3. First let us consider the collision between 'A' and 'B'
• Since all balls are identical, we can write mA = mB = mC = m
• If there is no external force acting on the system, the law of conservation of momentum is valid
• So we have:
$\mathbf\small{mv+m \times 0=m v_{Af}+m v_{Bf}}$
• Substituting the values, we get:
$\mathbf\small{\Rightarrow v= v_{Af}+ v_{Bf}}$
4. In an elastic collision, kinetic energy is conserved. So we have:
$\mathbf\small{m_A v_{Ai}^2+m_B v_{Bi}^2=m_A v_{Af}^2+m_B v_{Bf}^2}$
• Substituting the values, we get:
$\mathbf\small{mv^2+m \times 0^2=m v_{Af}^2+m v_{Bf}^2}$
$\mathbf\small{\Rightarrow v^2=v_{Af}^2+v_{Bf}^2}$
5. Squaring the result in (3), we get: $\mathbf\small{v^2= v_{Af}^2+2v_{Af}\;v_{Bf}+ v_{Bf}^2}$
Substituting for $\mathbf\small{v^2}$ from(4), we get: $\mathbf\small{v_{Af}^2+ v_{Bf}^2= v_{Af}^2+2v_{Af}\;v_{Bf}+ v_{Bf}^2}$
$\mathbf\small{\Rightarrow 0= 2v_{Af}\;v_{Bf}}$
• Final velocity of B (vBf) cannot be zero because, it is hit by the ball 'A' of equal mass
• And also, there is no friction to stop 'B'
• So we get vAf = 0
6. Substituting this value of vAf in (3), we get: vBf = v
• So we can write a summary about the collision between A and B:
(i) Two balls 'A' and 'B' collide elastically with each other
(ii) The two masses are equal. That is., mA = mB = m
(iii) Before the collision, 'A' was moving with a velocity v
(iv) Before the collision, 'B' was stationary. That is., vBi = 0
(v) After the collision, 'A' becomes stationary
(vi) After the collision, 'B' begins to move with v (the same velocity with which 'A' was moving)
7. Now we consider the collision between 'B' and 'C'
• B has an initial velocity of 'v' and C is stationary
• So this is similar to the collision between A and B
• The steps will be same as from (3) to (6) above
• We can write a summary about the collision between B and C:
(i) Two balls 'B' and 'C' collide elastically with each other
(ii) The two masses are equal. That is., mB = mC = m
(iii) Before the collision, 'B' was moving with a velocity v
(iv) Before the collision, 'C' was stationary. That is., vCi = 0
(v) After the collision, 'B' becomes stationary
(vi) After the collision, 'C' begins to move with v (the same velocity with which 'B' was moving)
8. So we see that, out of the 3 options given on the right side in fig.6.71 above, (ii) is the correct option
Easy method:
1. We have derived formulas for vAf and vBf in the case of elastic one-dimensional collision between two objects A and B (with B initially at rest)
• The formulas are:
$\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
$\mathbf\small{v_{Bf}=\frac{2m_Av_{Ai}}{(m_A + m_B)}}$
(Details here)
2. Further, we saw special case 1 in which the two objects have the same mass
• The collision between balls A and B in this problem, will fall under this case 1
• So we can directly write the result:
After the collision, 'A' will come to a stop and 'B' will begin to move with velocity 'v'
3. The collision between 'B' and 'C' also falls under this case 1
• So we can directly write the result:
• After the collision, 'B' will come to a stop and 'C' will begin to move with velocity 'v'
4. So we see that, out of the 3 options given on the right side in fig.6.71 above, (ii) is the correct option
Solved example 6.38
The bob A of a pendulum released from 30o to the vertical hits another bob B of the same mass at rest on a table as shown in fig. 6.72(a). How high does the bob A rise after the collision? Neglect the size of the bobs and assume the collision to be elastic
 |
| Fig.6.72 |
1. The bob B is kept vertically below the point O
• The bob B is on the path of A. This is shown in fig.b
• So when the bob A hits B, it will be a one-dimensional collision
2. Given that, the two bobs have the same mass
3. Given that the collision is elastic
4. Given that, the size of the bobs can be neglected
• So we can assume that they have the same diameter
5. Given that B is at rest
6. Based on the five points above, the collision falls under special case 1
• We have: $\mathbf\small{v_{Af}=\frac{(m_A-m_B)v_{Ai}}{(m_A + m_B)}}$
• In special case 1, mA = mB
• This gives vAf = 0
• So we can directly write the result:
■ After the collision, 'A' will come to a stop and 'B' will begin to move with velocity 'v'
♦ Where 'v' is the velocity attained by the bob A when it reaches the lowest point
7. Thus we see that, the bob A will not rise to any height. It will come to a stop at the very instant when collision occurs
• We will get the same result even if we change the angle 30o to any other value
While performing a collision experiment on a horizontal surface, we face three difficulties. They are:
1. There will be friction between the colliding bodies and the surface
• So some energy will be lost for doing work against friction
• As a result, their will be retardation (negative acceleration) and so the bodies will not be traveling with uniform velocities
• So we cannot write the conservation of kinetic energy equation
2. If the colliding bodies have different sizes, the velocities will not fall along the same straight line. So we will not get a one-dimensional collision
• This is shown in fig.6.73 below:
 |
| Fig.6.73 |
If we do the collision experiment in a vertical direction, all the 3 difficulties can be over come. Let us see how it is done:
• Let us take a basketball. It has a mass mB = 0.58 kg
2. Allow it to fall freely from a convenient height (h1) of say 1.5 m. This is shown in fig.6.74(a) below:
 |
| Fig.6.74 |
3. We want the velocity (v1) with which it hits the ground
• That is., we want the velocity just before collision
• We can easily obtain it using the familiar equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
$\mathbf\small{v_1^2=2\times g \times h_1=2\times 9.8 \times 1.5=29.4}$
$\mathbf\small{\therefore v_1=\sqrt{29.4}=5.422\;\;\text{ms}^{-1} }$
4. We want the velocity (v2) with which it leaves the ground after rebound
• That is., we want the velocity just after collision
• We can easily obtain this also using the same equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
• $\mathbf\small{v_2^2=2\times g \times h_2=2\times 10 \times 1.2=23.52}$
• $\mathbf\small{\therefore v_2=\sqrt{23.52}=4.849\;\;\text{ms}^{-1} }$
5. Once we determine v1 and v2, we can obtain the coefficient of restitution using the relation:
$\mathbf\small{\text{Coefficient of restitution}\;\;(e)=\frac{\text{Velocity after collision}}{\text{Velocity before collision}}=\frac{v_2}{v_1}=\frac{4.849}{5.422}=0.894}$
6. Now take a smaller tennis ball/rubber ball/ table tennis ball
• Let us take a tennis ball. It has a mass mT = 0.057 kg
7. While holding the basketball, stack the tennis ball vertically above it
• Drop them together from the same height h1 of 1.5 m. This is shown in fig.6.74(c)
• Following two points must be ensured while making the drop:
(i) The two balls must move together
(ii) The smaller ball must be exactly vertical above the larger ball
• Then only we will get a one-dimensional collision
8. Measure the height (h3) upto which the basketball rebounds
Let h3 = 0.75 m. This is shown in fig.d
9. After collision with the ground, we find that:
• The tennis ball rebounds to a larger height (h4). This is shown in fig.e
• It is not easy to measure that h4
• So we will use analytical method to find h4
• While using the analytical method, we will see the applications of:
♦ Conservation of momentum
♦ Conservation of energy
10. Note that, the basketball does not rebound to the same height obtained earlier
That is., h3 < h2
11. Let us analyse what happens during the collision:
• There are actually two collisions
(i) The first one is that between the basketball and the floor
(ii) As a result of that collision, the basket ball rebounds and collides with the tennis ball which was trailing close behind. This is the second collision
■ This second collision is the one which is of interest to us
12. For this second collision, we have:
• mB = 0.58 kg, mT = 0.057 kg
• vBi = v2 = 4.849 ms-1
[∵ the basket ball will have a rebound velocity of 4.849 ms-1 after hitting the floor. This we obtained in (4)]
• vTi = v1 = -5.422 ms-1
[∵ any mass falling through 1.5 m, will have a velocity of 5.422 ms-1. This we obtained in (3)
The negative sign is given because it is in the downward direction]
• vBf = ?, vTf = ?
13. Let us apply the principle of conservation of momentum:
$\mathbf\small{m_B\,v_{Bi}+m_T\,v_{Ti}=m_B\,v_{Bf}+m_T\,v_{Tf}}$
• Substituting the known values, we get:
$\mathbf\small{0.58\times 4.849-0.057\times 5.422=0.58\,v_{Bf}+0.057\,v_{Tf}}$
$\mathbf\small{\Rightarrow 0.58\,v_{Bf}+0.057\,v_{Tf}=2.503}$
$\mathbf\small{\Rightarrow 10.175\,v_{Bf}+\,v_{Tf}=43.926}$
$\mathbf\small{\Rightarrow v_{Tf}=43.926-10.175\,v_{Bf}}$
14. Let us apply the principle of conservation of kinetic energy:
• Assuming an elastic collision, we have:
$\mathbf\small{\frac{1}{2}m_B\,v_{Bi}^2+\frac{1}{2}m_T\,(-v_{Ti})^2=\frac{1}{2}m_B\,v_{Bf}^2+\frac{1}{2}m_T\,v_{Tf}^2}$
• Multiplying both sides by '2', we get:
$\mathbf\small{m_B\,v_{Bi}^2+m_T\,v_{Ti}^2=m_B\,v_{Bf}^2+m_T\,v_{Tf}^2}$
• Substituting the known values, we get:
$\mathbf\small{13.642+1.676=0.58\,v_{Bf}^2+0.057\,v_{Tf}^2}$
$\mathbf\small{\Rightarrow 15.317=0.58\,v_{Bf}^2+0.057\,v_{Tf}^2}$
15. So we have two equations:
(i) From (13) we have: $\mathbf\small{v_{Tf}=43.926-10.175\,v_{Bf}}$
(ii) From (14) we have: $\mathbf\small{0.58\,v_{Bf}^2+0.057\,v_{Tf}^2=15.317}$
• Solving them, we get: vBf = 4.853 OR 3.01
• When vBf = 4.853, we get vTf = -5.453 ms-1
• When vBf = 3.01, we get vBf = 13.3 ms-1
16. The negative velocity is not acceptable because, the tennis ball moves upwards after collision
■ So we can write:
After the collision, the tennis ball moves upwards with the velocity vTf = 13.3 ms-1
17. The height upto which the tennis ball reaches can be calculated using the familiar equation for free fall: $\mathbf\small{v^2=2gh}$
• Substituting the values we get:
$\mathbf\small{v_{Tf}^2=2\times g \times h_4}$
$\mathbf\small{\Rightarrow 13.3^2=2\times 9.8 \times h_4}$
$\mathbf\small{\Rightarrow h_4=9.025\;\text{m}}$
■ 13.3 ms-1 is a very high velocity. And 9.025 m is nearly 3 storeys high. So all necessary safety precautions should be adopted while performing this experiment
In the next section, we will discuss about two-dimensional collisions
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