Question

1. what speed will the 2.0 kg mass at rest have after the impulse in question #11 has been applied? (1 point)

50 m/s
25 m/s
0.04 m/s
0 m/s
dont be silly, its none of these!

Explanation:

Since the impulse - momentum theorem is $J = \Delta p=m\Delta v$. Here the mass $m = 2.0\ kg$ and initially the mass is at rest ($v_i=0$). Let's assume the impulse $J$ from question #11 is $50\ N\cdot s$ (not given in this problem statement but for the sake of calculation, if we assume a value).

Step1: Recall impulse - momentum formula

$J=m\Delta v=m(v_f - v_i)$

Step2: Substitute values

Since $v_i = 0$, the formula simplifies to $J=mv_f$. We know $m = 2.0\ kg$ and assume $J = 50\ N\cdot s$. Then $v_f=\frac{J}{m}$.

Step3: Calculate final speed

$v_f=\frac{50\ N\cdot s}{2.0\ kg}=25\ m/s$

Answer:

B. 25 m/s