# Impulse imparted during elastic collision [closed]

Lets say there are two objects a of 1kg and a very heavy block. The light mass has a velocity of $$12$$ $$ms^{-1}$$ towards right. The heavy block has a velocity of $$10$$ $$ms^{-1}$$ towards right. The light mass collides elastically with the heavy block. What is the impulse imparted by the heavy block? I am facing a problem calculating change in velocity of ball on hitting the moving block which is required to calculate impulse.

My Attempt: As this is an elastic collision

Coefficient of restitution $$=1$$

Therefore,

$$1=\frac{12-10}{10-v_{ball}}$$

Which gives final velocity of ball to be $$8$$ $$ms^{-1}$$ towards right.

Calculating change in momentum of ball that is momentum imparted to ball during collision:

Initial velocity of ball =$$12$$ $$ms^{-1}$$ towards right.

Final velocity of ball = $$8$$ $$ms^{-1}$$ towards right.

Velocity of approach of ball=$$(12-10)=2$$ $$ms^{-1}$$ towards right.

In the problem below velocity with which water hits the plate is taken as velocity of approach

So while calculating impulse imparted to ball(1kg) by the block, Should we take

$$1.$$ m(final velocity - velocity with which ball hits the block)

$$1(8-2)=6$$ $$kg$$ $$ms^{-1}$$ towards right.

$$2.$$ m(final velocity - initial velocity)

$$1(8-12)=4$$ $$kg$$ $$ms^{-1}$$ towards left.

I think that the $$(1)$$ can't be true as block would apply a pushing force towards left thus imparting impulse towards left. Which means $$(2)$$ should be correct. Am I wrong?

In such a case where $$(2)$$ is correct why was velocity of approach used to calculate force on wall in the water jet problem and why it is not applicable in block and ball case. I am facing a problem in calculating change in velocity of ball when it hits the moving block.

## closed as off-topic by John Rennie, stafusa, Kyle Kanos, Jon Custer, ZeroTheHeroMar 2 at 3:21

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Your confusion comes from mixing two different frames of reference. In the frame of reference where the ball initially travels at $$12 \frac{m}{s}$$, it's final velocity is $$8 \frac{m}{s}$$ and the impulse is $$m\Delta v = 1 kg (8 - 12) \frac{m}{s} = -4 \frac{kg \cdot m}{s}$$ (or $$4 \frac{kg \cdot m}{s}$$ to the left as you said). When you calculate using the "velocity of approach" of the ball, you're really just in a frame of reference where the large mass is taken to be stationary so that all velocity is relative to the mass. The velocity of an object in this frame in terms of the old one is given by $$v_{new} = v_{old} - 10 \frac{m}{s}$$ In this frame, the ball is moving at $$(12 - 10) \frac{m}{s} = 2 \frac{m}{s}$$ to the right. The velocity of the ball when it rebounds is no longer $$8 \frac{m}{s}$$, that was in the old frame of reference. In this frame of reference, it is $$(8 - 10) \frac{m}{s} = -2 \frac{m}{s}$$, so the impulse is $$1 kg (-2 - 2) \frac{m}{s} = -4 \frac{kg \cdot m}{s}$$ same as the original answer. You can do calculations in any frame of reference, you just have to be consistent and not switch around in the middle.