# A 45kg cart traveling at 24 m/s collides with a stationary 28 kg cart. a) after the collision, the 45 kg cart continues forward with a speed of 12 m/s. what is the speed of the cart after this collision? b) calculate the impulse delivered to the 45 kg cart.

(A) Momentum must be conserved in the collision, so the final total momentum of both carts must be the same as the initial momentum of the first cart:
`m_1 v_{1i} + m_2 v_{2i} = m_1 v_{1f} + m_2 v_{2f}`
`(45 kg)(24 m/s) + (0) = (45 kg)(12 m/s) + (28 kg) v_{2f}`
One equation, one unknown, so we can solve.
`540 kg*m/s = (28 kg) v_{2f}`
`v_{2f} = 19.29 m/s`
Everything is given in two significant figures, so this is our final answer for the velocity:
`v_{2f} = 19 m/s`

(B) The impulse delivered to the second cart is just the difference between its initial momentum (zero) and its final momentum, so it is just the final momentum, which is 540.12 kg*m/s, or 540 kg*m/s in two significant figures.

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