Question

for the given pre - and post - collision information, identify the collision parameters that combine to produce the indicated momentum change. pick two sets of parameters. initial state v = 6 m/s 2 kg collision parameters final state v = 14 m/s 2 kg tap an option below to select or deselect an answer. if two options are already selected, first deselect one. (a + sign indicates a rightward direction; a - sign indicates a leftward direction) δp = +24 kg·m/s f = + 32 n, δt = 0.5 s f = +8 n, δt = 2 s impulse = -8 n·s impulse = +8 n·s

Explanation:

Step1: Recall impulse - momentum theorem

The impulse - momentum theorem states that $\Delta p = F\Delta t$, where $\Delta p$ is the change in momentum, $F$ is the average force, and $\Delta t$ is the time interval.

Step2: Calculate impulse for each force - time pair

For $F = + 32\ N$ and $\Delta t=0.5\ s$, the impulse $J_1=F\Delta t=(+32\ N)\times(0.5\ s)= + 16\ N\cdot s$.
For $F = + 8\ N$ and $\Delta t = 2\ s$, the impulse $J_2=F\Delta t=(+8\ N)\times(2\ s)=+16\ N\cdot s$.
The given $\Delta p=+24\ kg\cdot m/s = + 24\ N\cdot s$. The impulse values we calculated do not match the given $\Delta p$. But we know that impulse is equal to the change in momentum. The correct impulses for the given $\Delta p = + 24\ kg\cdot m/s$ are the ones that satisfy $\Delta p=J$.
We are given two impulse values: impulse $=- 8\ N\cdot s$ and impulse $=+8\ N\cdot s$. These are not correct for the given $\Delta p = + 24\ kg\cdot m/s$. However, if we consider the force - time pairs:
For $F = + 32\ N$ and $\Delta t = 0.5\ s$, $J = F\Delta t=32\times0.5=16\ N\cdot s$ (wrong).
For $F = + 8\ N$ and $\Delta t=2\ s$, $J = F\Delta t=8\times2 = 16\ N\cdot s$ (wrong).
Let's use the correct relationship. We know that $\Delta p=J$.
If we consider the force - time combinations:

1. For $F = + 32\ N$ and $\Delta t = 0.75\ s$, $J=F\Delta t=32\times0.75 = 24\ N\cdot s$.
2. Another way is to use the impulse directly. Since $\Delta p = J$, and we want $\Delta p=+24\ kg\cdot m/s=+24\ N\cdot s$.

We know that impulse $J = F\Delta t$. If we assume $F = + 12\ N$ and $\Delta t = 2\ s$, $J=F\Delta t=(+12\ N)\times(2\ s)=+24\ N\cdot s$

The two sets of parameters that can produce $\Delta p=+24\ kg\cdot m/s$ are:
Set 1: $F = + 12\ N$, $\Delta t=2\ s$ (because $J = F\Delta t=(+12\ N)\times(2\ s)=+24\ N\cdot s$)
Set 2: If we consider the impulse directly, an impulse of $+24\ N\cdot s$ (since $\Delta p = J$)

Answer:

Set 1: $F = + 12\ N$, $\Delta t = 2\ s$; Set 2: Impulse $=+24\ N\cdot s$