Question

A 7kg bowling ball collided head on with a stationary 2kg bowling pin. The pin flies forward with a velocity of 11.04 m/s. If the ball continues forward with a speed of 1.8 m/s what was the initial velocity of the ball?

Answer 1

We will have the following:

We are given the following:

$\begin{cases}m_{\text{ball}}=9\operatorname{kg} \\ m_{pin}=2\operatorname{kg} \\ v_{pin}=11.04m/s \\ v_{\text{ball}}=1.8m/s\end{cases}$

So, we will have to proceed as follows:

First, we calculate the momentum of the pin:

$\tau_{pin}=(2\operatorname{kg})(11.04m/s)\Rightarrow\tau_{pin}=22.08\operatorname{kg}\cdot m/s$

Now, we calculate the final momentum of the ball:

$\tau_{f\text{ball}}=(7\operatorname{kg})(1.8m/s^2)\Rightarrow\tau_{\text{fball}}=12.6\operatorname{kg}\cdot m/s$

Now, we will have that the original momentum of the ball would be:

$\tau_{\text{iball}}=\tau_{\text{fball}}+\tau_{pin}\Rightarrow\tau_{\text{iball}}=12.6\operatorname{kg}\cdot m/s+22.08\operatorname{kg}\cdot m/s$$\Rightarrow\tau_{\text{iball}}=34.68\operatorname{kg}\cdot m/s$

Now, we find it's initial speed:

$\tau_{\text{iball}}=m\cdot v_i\Rightarrow34.68\operatorname{kg}\cdot m/s=(7\operatorname{kg})\cdot v_i$$\Rightarrow v_i=\frac{867}{175}m/s\Rightarrow v_i\approx4.95m/s$

So, the initial speed of the ball was of 867/175 m/s, that is approximately 4.95 m/s.