Question

pre - and post - collision information is shown. identify the collision parameters that are consistent with the indicated change in momentum. initial state v = 10 m/s 8 kg collision parameters final state v = 6 m/s 8 kg tap an option below to select or change an answer. (a + sign indicates a rightward direction; a - sign indicates a leftward direction.) impulse = -64 n·s f = -16 n, δt = 2 s f = -16 n, δt = 1 s δp = -16 kg·m/s δp = -4 kg·m/s

Explanation:

Step1: Recall the formula for momentum

Momentum $p = mv$. Initial momentum $p_i=mv_i$, where $m = 8$ kg and $v_i = 10$ m/s, so $p_i=8\times10=80$ kg·m/s. Final momentum $p_f=mv_f$, with $m = 8$ kg and $v_f = 6$ m/s, so $p_f=8\times6 = 48$ kg·m/s.

Step2: Calculate the change in momentum

The change in momentum $\Delta p=p_f - p_i$. So $\Delta p=48 - 80=- 32$ kg·m/s. Also, recall the impulse - momentum theorem $J=\Delta p=F\Delta t$.

Step3: Check each option

For impulse $J=-64$ N·s, this is not equal to $\Delta p=-32$ kg·m/s. For $F=-16$ N and $\Delta t = 2$ s, $J=F\Delta t=-16\times2=-32$ N·s = - 32 kg·m/s (since 1 N·s=1 kg·m/s). For $F=-16$ N and $\Delta t = 1$ s, $J=F\Delta t=-16\times1=-16$ N·s. $\Delta p=-16$ kg·m/s and $\Delta p=-4$ kg·m/s do not match the calculated $\Delta p=-32$ kg·m/s.

Answer:

$F=-16$ N, $\Delta t = 2$ s