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)=(12)/(1.5) \\ v_(f2)=8 \end{gathered}`

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