How Is Impulse Calculated in a Two-Cart Collision?

[dylanhouse]

A 1.0 kg cart traveling at 1.0m/s right hits a 4.0kg cart at rest. After the collision, the lighter cart is observed to move to the left at 0.5m/s. What impulse did the interaction deliver to the massive cart (magnitude and direction)? What is the carts velocity after the collision?

I calculated an impulse of 1.5 kg m/s right, though I'm not sure if this is correct. I'm not sure of the procedure for this question.

 

[TysonM8]

Kinetic energy is conserved in perfectly elastic collisions.

Initial KE of 1kg cart = 0.5 x 1 x 1^2
= 0.5 joules

Final KE of 1kg cart = 0.5 x 1 x 0.5^2
= 0.125 joules

As kinetic energy is conserved, KE of the 4kg cart is equal to 0.5-0.125
= 0.375 joules

0.375 = 0.5 x 4 x v^2
v^2 = 3
v = sqr(3) m/s

Impulse = change in momentum

Initial momentum = 0 as the cart is at rest.

Final momentum = mv
= 4sqr(3) kg m/s

The impulse exerted on the large cart = 4sqr(3) Ns to the right and produced a velocity of sqr(3) m/s also to the right.

 

[ehild]

A 1.0 kg cart traveling at 1.0m/s right hits a 4.0kg cart at rest. After the collision, the lighter cart is observed to move to the left at 0.5m/s. What impulse did the interaction deliver to the massive cart (magnitude and direction)? What is the carts velocity after the collision?

I calculated an impulse of 1.5 kg m/s right, though I'm not sure if this is correct. I'm not sure of the procedure for this question.

The change of momentum of a body is equal to the impulse it gained in the interaction. The change is momentum of the light cart is m(v2-v1), (negative) and it delivers impulse of the same magnitude, but positive for the more massive cart. (Newton's Third Law) The total momentum is conserved when there is no external force!

You got the correct magnitude of the impulse. It is equal to the change of momentum of the heavier cart. You can determine the change of velocity from that.


ehild

 

[ehild]

Kinetic energy is conserved in perfectly elastic collisions.

According to the given data, it is not a perfectly elastic collision. You can not use conservation of energy. But conservation of momentum holds for every collision.


ehild