Answer to Question #286866 in Mechanics | Relativity for David

Question

A bullet, of mass 55.0g is fired at 251m/s toward a block of mass 7.50kg that is sitting on the floor. The bullet becomes embedded in the block and the block and bullet combination slides 0.755m along the floor before coming to rest. Determine the coefficient of kinetic friction between the block and the floor.

Answer 1

Explanations & Calculations

• First, the speed of the combination needs to be found using conservation of linear momentum.

$\qquad\qquad \begin{aligned} \small m_{bullet}.v_b+m_{Block}.v_B&=\small (m_b+m_B).V_{Bb}\\ \small V_{Bb}&=\small \frac{0.055\,kg\times251\,ms^{-1}+0}{(0.055+7.50)kg}\\ &=\small 1.83\,ms^{-1} \end{aligned}$

• Once it is found, the distance travelled can be assessed using the conservation of mechanical energy along with non-conservative forces.

$\qquad\qquad \begin{aligned} \small \frac{1}{2}(m_B+m_b)V_{Bb}^2&=\small fs\\ \small \frac{1}{2}(m_B+m_b)V_{Bb}^2&=\small \mu_k.(m_B+m_b)g.s\\ \small \mu_k&=\small \frac{V_{Bb}^2}{2gs} \end{aligned}$

• You now can substitute the known and calculated values into this & try getting the answer.

• Overall what happens here is that, upon impact, the system obtains a velocity to move there onwards.
• As it moves ahead, the friction (kinetic friction) from the ground, opposes the motion absorbing its kinetic energy thereby taking the system to a stop.
• For that energy conservation can be used.