- #1

- 42

- 0

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.

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dylanhouse
- Start date

- #1

- 42

- 0

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.

- #2

- 25

- 1

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.

- #3

ehild

Homework Helper

- 15,543

- 1,912

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

- #4

ehild

Homework Helper

- 15,543

- 1,912

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

Share: