In the previous section we saw Newton's second law of motion. In this section we will learn about Impulse. We will write it in steps:
1. Consider a batsman hitting hard on a ball with his bat.
• The time duration of contact between the bat and the ball will be very small.
• Yet, a large force is applied on the ball.
2. We want to know how much force is applied on the ball
• For that, we want the following items:
(i) Mass of the ball (m)
(ii) Initial velocity of the ball (v1)
(iii) Final velocity of the ball (v2)
(iv) Time duration in which the 'change from v1 to v2 happens' (Δt)
♦ That is., the time duration for which the force acts
3. So, from v1, v2 and Δt, we find a
• For that, we can use the equation: $\mathbf\small{a=\frac{(v_2-v_1)}{\Delta t}}$
• The product of m and a should give the force applied on the ball by the batsman
4. But since the time duration (Δt) is very small, we are not able to measure it.
♦ That is., ‘Δt’ in this case is not a measurable quantity.
• Consequently, we cannot calculate $\mathbf\small{\frac{(v_2-v_1)}{\Delta t}}$
♦ That is., we cannot calculate a
• So we cannot find the product of mass and acceleration
5. In such cases, we follow another method to obtain the ‘effect on the ball’
• Consider the familiar equation: $\mathbf\small{F=\frac{(mv_2-mv_1)}{\Delta t}}$
• Bring Δt to the left side. We get: F Δt = (mv2-mv1)
■ We cannot calculate the left side of the above equation because, Δt is very small and hence not measurable
6. But what about the right side?
• We can measure the initial velocity v1 and the final velocity v2
■ So we can calculate the right side.
• We know that the right side is in fact, the change in momentum experienced by the ball
7. So we can write the two points below:
(i) Consider the product of the following two items:
(a) Force experienced by the ball (F)
(b)The time for which F is experienced by the ball (Δt)
(ii) The product of the two items ‘F×Δt’ will be equal to ‘the change in momentum’ experienced by the ball
8. The product 'F×Δt' is called Impulse
• So we can write Eq.5.3: Impulse = F×Δt = m(v2-v1)
• Though time is a scalar quantity, force is a vector quantity. So impulse is a vector quantity
• The unit for impulse would be obviously: newton-second (N s)
A batsman hits back a ball straight in the direction of the bowler without changing it’s initial speed of 12 ms-1. If the mass of the ball is 0.15 kg, determine the impulse imparted on the ball. (Assume linear motion of the ball)
Solution:
1. The bowler bowls the ball towards the batsman
• So the initial direction is: bowler → batsman
2. Given that, the batsman hits back a ball straight in the direction of the bowler
• So after hitting, the direction is: batsman → bowler
■ Thus the final direction is the exact opposite of the initial direction
3. Given that, initial speed does not change.
• So we can write:
Magnitude of initial speed = magnitude of final speed = 12 ms-1
• Taking directions into account, we get:
v1 = 12 ms-1 (positive direction for: bowler → batsman)
v2 = -12 ms-1 (negative direction for the opposite: batsman → bowler)
4. Impulse imparted = F×Δt = m(v2-v1) = 0.15×(-12-12) = (0.15 × -24) = -3.6 N s
• The negative direction is for the direction: batsman → bowler
• So the impulse is directed from the batsman towards the bowler
Solved example 5.8
A batsman deflects a ball at an angle of 45o without changing it’s initial speed, which is equal to 54 km/h. What is the impulse imparted to the ball? Mass of the ball = 0.15 kg
Solution:
1. The bowler bowls the ball towards the batsman
• So the initial direction is: bowler → batsman.
• It is shown by the magenta arrow in fig.5.16(a) below:
2. Let this initial direction be parallel to the x axis
• The final direction is inclined at 45o to the initial direction
• It is shown by the cyan arrow
(Note that, the cyan arrow must point away from the batsman. It cannot point towards the batsman)
3. Let us split the problem into two parts:
(a) Impulse in the x direction
(b) Impulse in the y direction
4. So we have to split the velocities into x and y components:
(a) Magnitude of the initial velocity in the x direction = $\mathbf\small{|\vec{v_{1x}}|=|\vec{v_{1}}|=15\: ms^{-1}}$
[∵ 54 km/h = 15 ms-1]
• Magnitude of the final velocity in the x direction = $\mathbf\small{|\vec{v_{2x}}|=|\vec{v_{2}}|\cos 45=(-15 \cos 45) \: ms^{-1}}$
(The '-' sign is given because, it's direction is towards the negative side of the x axis)
(b) Magnitude of the initial velocity in the y direction = 0
• Magnitude of the final velocity in the y direction = $\mathbf\small{|\vec{v_{2y}}|=|\vec{v_{2}}|\sin 45=(15 \sin 45) \: ms^{-1}}$
5. Impulse in the x direction = 'Change in momentum' in the x direction
= $\mathbf\small{\vec{\Delta p_x}=(\vec{p_{2x}}-\vec{p_{1x}})=[m(|\vec{v_{2x}}|-|\vec{v_{1x}}|)]\hat{i}=[0.15(-15 \cos 45-15)]\hat{i}}$
= $\mathbf\small{[-0.15\times 15(\cos 45+1)]\hat{i}=-3.841\, \hat{i}}$
• Impulse in the y direction = 'Change in momentum' in the y direction
= $\mathbf\small{\vec{\Delta p_y}=(\vec{p_{2y}}-\vec{p_{1y}})=[m(|\vec{v_{2y}}|-|\vec{v_{1y}}|)]\hat{j}=[0.15(15 \sin 45-0)]\hat{j}=1.591\: \hat{j}}$
• The two 'change in momentums' are shown in fig.b
• Note that, $\mathbf\small{\vec{\Delta p_x}}$ is shown from left to right because we obtained a negative value for it
6. Now, the resultant impulse is the 'resultant of the two change in momentums'
• This resultant is 'the resultant change in momenum' $\mathbf\small{\vec{\Delta p}}$
• We have: $\mathbf\small{|\vec{\Delta p}|=\sqrt{|\vec{\Delta p_x}|^2+|\vec{\Delta p_y}|^2}=\sqrt{(-3.841)^2+(1.591)^2}=4.157}$
• So $\mathbf\small{\vec{\Delta p}}$ has a magnitude of 4.157 N s
7. Now we want it's direction
We have: $\mathbf\small{\theta = \tan^{-1}\left( \frac{|\vec{\Delta p_y}|}{|\vec{\Delta p_x}|}\right )=\tan^{-1}\left( \frac{1.591}{3.841} \right )=22.5001^o}$
• Note that 22.5 is half of 45
8. So we can write:
• The resultant impulse has a magnitude of 4.157 N s
• It has a direction which bisects the angle between the initial and final velocity vectors
1. Consider a batsman hitting hard on a ball with his bat.
• The time duration of contact between the bat and the ball will be very small.
• Yet, a large force is applied on the ball.
2. We want to know how much force is applied on the ball
• For that, we want the following items:
(i) Mass of the ball (m)
(ii) Initial velocity of the ball (v1)
(iii) Final velocity of the ball (v2)
(iv) Time duration in which the 'change from v1 to v2 happens' (Δt)
♦ That is., the time duration for which the force acts
3. So, from v1, v2 and Δt, we find a
• For that, we can use the equation: $\mathbf\small{a=\frac{(v_2-v_1)}{\Delta t}}$
• The product of m and a should give the force applied on the ball by the batsman
4. But since the time duration (Δt) is very small, we are not able to measure it.
♦ That is., ‘Δt’ in this case is not a measurable quantity.
• Consequently, we cannot calculate $\mathbf\small{\frac{(v_2-v_1)}{\Delta t}}$
♦ That is., we cannot calculate a
• So we cannot find the product of mass and acceleration
5. In such cases, we follow another method to obtain the ‘effect on the ball’
• Consider the familiar equation: $\mathbf\small{F=\frac{(mv_2-mv_1)}{\Delta t}}$
• Bring Δt to the left side. We get: F Δt = (mv2-mv1)
■ We cannot calculate the left side of the above equation because, Δt is very small and hence not measurable
6. But what about the right side?
• We can measure the initial velocity v1 and the final velocity v2
■ So we can calculate the right side.
• We know that the right side is in fact, the change in momentum experienced by the ball
7. So we can write the two points below:
(i) Consider the product of the following two items:
(a) Force experienced by the ball (F)
(b)The time for which F is experienced by the ball (Δt)
(ii) The product of the two items ‘F×Δt’ will be equal to ‘the change in momentum’ experienced by the ball
8. The product 'F×Δt' is called Impulse
• So we can write Eq.5.3: Impulse = F×Δt = m(v2-v1)
• Though time is a scalar quantity, force is a vector quantity. So impulse is a vector quantity
• The unit for impulse would be obviously: newton-second (N s)
Solved example 5.7
Solution:
1. The bowler bowls the ball towards the batsman
• So the initial direction is: bowler → batsman
2. Given that, the batsman hits back a ball straight in the direction of the bowler
• So after hitting, the direction is: batsman → bowler
■ Thus the final direction is the exact opposite of the initial direction
3. Given that, initial speed does not change.
• So we can write:
Magnitude of initial speed = magnitude of final speed = 12 ms-1
• Taking directions into account, we get:
v1 = 12 ms-1 (positive direction for: bowler → batsman)
v2 = -12 ms-1 (negative direction for the opposite: batsman → bowler)
4. Impulse imparted = F×Δt = m(v2-v1) = 0.15×(-12-12) = (0.15 × -24) = -3.6 N s
• The negative direction is for the direction: batsman → bowler
• So the impulse is directed from the batsman towards the bowler
Solved example 5.8
A batsman deflects a ball at an angle of 45o without changing it’s initial speed, which is equal to 54 km/h. What is the impulse imparted to the ball? Mass of the ball = 0.15 kg
Solution:
1. The bowler bowls the ball towards the batsman
• So the initial direction is: bowler → batsman.
• It is shown by the magenta arrow in fig.5.16(a) below:
2. Let this initial direction be parallel to the x axis
• The final direction is inclined at 45o to the initial direction
• It is shown by the cyan arrow
(Note that, the cyan arrow must point away from the batsman. It cannot point towards the batsman)
3. Let us split the problem into two parts:
(a) Impulse in the x direction
(b) Impulse in the y direction
4. So we have to split the velocities into x and y components:
(a) Magnitude of the initial velocity in the x direction = $\mathbf\small{|\vec{v_{1x}}|=|\vec{v_{1}}|=15\: ms^{-1}}$
[∵ 54 km/h = 15 ms-1]
• Magnitude of the final velocity in the x direction = $\mathbf\small{|\vec{v_{2x}}|=|\vec{v_{2}}|\cos 45=(-15 \cos 45) \: ms^{-1}}$
(The '-' sign is given because, it's direction is towards the negative side of the x axis)
(b) Magnitude of the initial velocity in the y direction = 0
• Magnitude of the final velocity in the y direction = $\mathbf\small{|\vec{v_{2y}}|=|\vec{v_{2}}|\sin 45=(15 \sin 45) \: ms^{-1}}$
5. Impulse in the x direction = 'Change in momentum' in the x direction
= $\mathbf\small{\vec{\Delta p_x}=(\vec{p_{2x}}-\vec{p_{1x}})=[m(|\vec{v_{2x}}|-|\vec{v_{1x}}|)]\hat{i}=[0.15(-15 \cos 45-15)]\hat{i}}$
= $\mathbf\small{[-0.15\times 15(\cos 45+1)]\hat{i}=-3.841\, \hat{i}}$
• Impulse in the y direction = 'Change in momentum' in the y direction
= $\mathbf\small{\vec{\Delta p_y}=(\vec{p_{2y}}-\vec{p_{1y}})=[m(|\vec{v_{2y}}|-|\vec{v_{1y}}|)]\hat{j}=[0.15(15 \sin 45-0)]\hat{j}=1.591\: \hat{j}}$
• The two 'change in momentums' are shown in fig.b
• Note that, $\mathbf\small{\vec{\Delta p_x}}$ is shown from left to right because we obtained a negative value for it
6. Now, the resultant impulse is the 'resultant of the two change in momentums'
• This resultant is 'the resultant change in momenum' $\mathbf\small{\vec{\Delta p}}$
• We have: $\mathbf\small{|\vec{\Delta p}|=\sqrt{|\vec{\Delta p_x}|^2+|\vec{\Delta p_y}|^2}=\sqrt{(-3.841)^2+(1.591)^2}=4.157}$
• So $\mathbf\small{\vec{\Delta p}}$ has a magnitude of 4.157 N s
7. Now we want it's direction
We have: $\mathbf\small{\theta = \tan^{-1}\left( \frac{|\vec{\Delta p_y}|}{|\vec{\Delta p_x}|}\right )=\tan^{-1}\left( \frac{1.591}{3.841} \right )=22.5001^o}$
• Note that 22.5 is half of 45
8. So we can write:
• The resultant impulse has a magnitude of 4.157 N s
• It has a direction which bisects the angle between the initial and final velocity vectors
In the next section we will see Newtons' third law
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