Question

A 3.20-kg block of wood rests on the muzzle opening of a vertically oriented rifle, the stock of the rifle being firmly planted on the ground. When the rifle is fired, an 8.10-g bullet (velocity = 7.50 multiplied by 102 m/s, straight upward) becomes completely embedded in the block.(a) Using the conservation of linear momentum, find the velocity of the block/bullet system immediately after the collision.(b) Ignoring air resistance, determine how high the block/bullet system rises above the muzzle opening of the rifle. A 3.20-kg block of wood rests on the muzzle opening of a vertically oriented rifle, the stock of the rifle being firmly planted on the ground. When the rifle is fired, an 8.10-g bullet (velocity = 7.50 multiplied by 102 m/s, straight upward) becomes completely embedded in the block.(a) Using the conservation of linear momentum, find the velocity of the block/bullet system immediately after the collision.(b) Ignoring air resistance, determine how high the block/bullet system rises above the muzzle opening of the rifle. @Physics

Answer 1

Is the velocity 7.50x102 or 7.50x10^2? Assuming the latter...

Solving for velocity based on momentum:
\[\Large \begin{array}{l}
p = mv\\
{p_i} = {p_f}\\
{m_i}{v_i} = {m_f}{v_f}\\
{\rm{0}}{\rm{.0081}} \cdot 7.50 \times {10^2} = {\rm{3}}{\rm{.2081}} \cdot {v_2}\\
{p_i} = 6.075 = {p_f} = 3.2081 \cdot {v_2}\\
{v_2} = 6.075/3.2081 = 1.894\,{\rm{m}}{{\rm{s}}^{ - 1}}
\end{array}\]So the block and bullet system have a velocity of about 1.894 m/s after impact.

To determine how far the system travels, use the kinematic equations, remembering that the only acceleration is gravity (-9.81m/s^2):
\[\Large \begin{array}{l}
v = u + at\\
t = \frac{{v - u}}{a} = \frac{{0 - 1.894}}{{ - 9.81}} = 0.193{\rm{ s}}\\
s = ut + \frac{1}{2}a{t^2} = 1.894 \cdot 0.193 - \frac{1}{2}9.81 \cdot {0.193^2} = 1.904{\rm{ m}}
\end{array}\]
So the system should travel about 1.904m.