Question

Thursday, February 11, 202110:08 AMCSA bullet travelling 420 m/s strikesa 1.5 kg target. How fast is thetarget moving if the bullet has amass of 0.1kg and is travelling300 m/s after the collision.

Answer 1

To solve this problem, we have to use conservation of momentum, which states that the initial momentum is equal to the final momentum. We know that momentum is defined as

$p=mv$

Let's apply it to the problem

$\begin{gathered} p_{i1}+p_{i2}=p_{f1}+p_{f2} \\ m_1\cdot v_{i1}+m_2\cdot v_{i2}=m_1\cdot v_{f1}+m_2\cdot v_{f2} \end{gathered}$

Using all the given information, we have

$0.1\cdot420+1.5\cdot0=0.1\cdot300+1.5\cdot v_{f2}$

Now, we solve the equation for v_f2:

$\begin{gathered} 42=30+1.5\cdot v_{f2} \\ 42-30=1.5\cdot v_{f2} \\ 12=1.5\cdot v_{f2} \\ v_{f2}=\frac{12}{1.5} \\ v_{f2}=8 \end{gathered}$

Hence, the target is moving 8 m/s after the collision.