Question

The 10-kg block is held at rest on the smooth inclined plane by the stop block at A. If the 10-g bullet is traveling at 300 m/s when it becomes embedded in the 10-kg block, determine the distance the block will slide up along the plane before momentarily stopping.

Answer 1

Answer:

$d=6.874mm$

Explanation:

Linear momentum of the block is conserved in the $x\prime$ direction since the impulsive force due to impact cancels each other internally.

$m_bv_b_x=(m_b+m_B)v_x\\(0.01)(300cos 30\textdegree)=(10+0.01)v\\\therefore v=0.259548m/s$

Using the conservation of energy to the block system considering the initial position:

$T_1+V_1=T_2+V_2\\\\0+0.5(m_b+m_B)v^2=0+(m_b+m_B)gh\\0.5(10+0.01)(0.259548)^2=(0.01+10)(9.8)\\h=0.003437m\\\therefore h=3.437mm$

Applying Sine rule:

$d=\frac{h}{Sin 30\textdegree}\\d=6.874mm$

Hence, the distance the block will slide up is 6.874mm