What is the impulse exerted on objects A and B during their collision?

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SUMMARY

The discussion centers on the collision between two steel balls, A and B, with masses of 0.6 kg and 0.2 kg, respectively. Before the collision, ball A moves at 8 m/s and ball B at 2 m/s. After the collision, the speed of ball B is twice that of ball A, which is calculated to be 4.4 m/s. The magnitude of the impulse exerted on ball B is determined to be 2.16 N·s, calculated using both the impulse exerted by A and B, confirming Newton's third law.

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  • Knowledge of basic physics equations: p=mv and Impulse = Ft = m(v-u)
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Homework Statement


Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving
towards each other in opposite directions on a smooth horizontal table when they collide
directly. Immediately before the collision, the speed of A is 8 m s–1 and the speed of B is 2 m s–1. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find
(a) the speed of A immediately after the collision,
(b) the magnitude of the impulse exerted on B in the collision.

Homework Equations


total momentum before collision = total momentum after collision
p=mv
Impulse = Ft = m(v-u)

The Attempt at a Solution


I've got the correct answer for (a), 4.4 m s-1 so I'll skip that bit. It's (b) that I don't get. It asks for the impulse exerted on B so that should be the impulse exerted by A

Ft = 0.6(4.4 - 8) = 0.6(-3.6) = -2.16 N s

since they ask for magnitude only

/therefore Impulse = 2.16 N s

But the answer given calculates the impulse exerted by B and still gets the same answer

Impulse = 0.2(2 + 8.8) = 2.16 N s

Why?

EDIT: Sorry don't know how to get the therefore symbol in LaTeX...my first time using it actually :D
 
Last edited:
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Hi Joe Jacobs,

Joe Jacobs said:

Homework Statement


Two small steel balls A and B have mass 0.6 kg and 0.2 kg respectively. They are moving
towards each other in opposite directions on a smooth horizontal table when they collide
directly. Immediately before the collision, the speed of A is 8 m s–1 and the speed of B is 2 m s–1. Immediately after the collision, the direction of motion of A is unchanged and the speed of B is twice the speed of A. Find
(a) the speed of A immediately after the collision,
(b) the magnitude of the impulse exerted on B in the collision.

Homework Equations


total momentum before collision = total momentum after collision
p=mv
Impulse = Ft = m(v-u)

The Attempt at a Solution


I've got the correct answer for (a), 4.4 m s-1 so I'll skip that bit. It's (b) that I don't get. It asks for the impulse exerted on B so that should be the impulse exerted by A

Ft = 0.6(4.4 - 8) = 0.6(-3.6) = -2.16 N s

This is the impulse exerted on A. The impulse on an object is equal to its change in momentum. Notice that this impulse is negative, and it is the ball B that puts a negative force on ball A. (Ball A felt a negative force, because it had a positive velocity and then slowed down.)

since they ask for magnitude only

/therefore Impulse = 2.16 N s

But the answer given calculates the impulse exerted by B and still gets the same answer

Impulse = 0.2(2 + 8.8) = 2.16 N s

Why?

If you are asking why the magnitudes are the same, think about what Newton's third law says about this situation.
 
This is the impulse exerted on A. The impulse on an object is equal to its change in momentum. Notice that this impulse is negative, and it is the ball B that puts a negative force on ball A. (Ball A felt a negative force, because it had a positive velocity and then slowed down.)
That was the answer I was looking for. Thx
 

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