Question

A rifle is aimed at a 2.0-kg block of wood along an inclined plane making an angle of 25°, as shown in Fig. 9–59. A 9.5-g bullet is fired at 760 m/s and becomes embedded in the block. How far up the incline does the block/bullet slide?
(a) Ignore the friction.
(b) Assume μₖ = 0.33.

Accepted Answer

Step 1: Analyze the problem and identify the key principles. This is a conservation of momentum and energy problem. First, calculate the velocity of the block and bullet system immediately after the collision using the principle of conservation of momentum. The initial momentum of the system is due to the bullet, and the final momentum is shared between the block and the embedded bullet. Step 2: Use the conservation of momentum formula: $$ m_b v_b = (m_b + m_w) v_f $$, where $$ m_b  is $$the mass of the bullet, $$ v_b  is $$the velocity of the bullet, $$ m_w  is $$the mass of the block, and $$ v_f  is $$the final velocity of the block and bullet system. Solve for $$ v_f . $$Step 3: For part (a), calculate the distance the block and bullet slide up the incline using the principle of conservation of energy. The initial kinetic energy of the block and bullet system is converted into gravitational potential energy as the system moves up the incline. Use the formula $$ KE = PE $$, where $$ \frac{1}{2}(m_b + m_w)v_f^2 = (m_b + m_w)g h . $$Solve for $$ h $$, the height, and then use trigonometry to find the distance along the incline: $$ d = \frac{h}{\sin 	heta} . $$Step 4: For part (b), include the effect of friction. The work done against friction reduces the distance the block and bullet slide. The frictional force is given by $$ F_f = \mu_k (m_b + m_w) g \cos 	heta . $$The work done by friction is $$ W_f = F_f d . $$Modify the energy conservation equation to include the work done by friction: $$ \frac{1}{2}(m_b + m_w)v_f^2 = (m_b + m_w)g h + W_f . $$Solve for $$ d $$ iteratively. Step 5: Summarize the process. For part (a), use conservation of energy without friction to find the distance. For part (b), account for friction by including the work done against friction in the energy equation. Ensure all calculations are consistent with the given values and units.

A rifle is aimed at a 2.0-kg block of wood along an inclined plane making an angle of 25°, as shown in Fig. 9–59. A 9.5-g bullet is fired at 760 m/s and becomes embedded in the block. How far up the incline does the block/bullet slide?
(a) Ignore the friction.
(b) Assume μₖ = 0.33.

Key Concepts

Conservation of Momentum

Kinetic Energy and Work-Energy Principle

Inclined Plane Dynamics

A rifle is aimed at a 2.0-kg block of wood along an inclined plane making an angle of 25°, as shown in Fig. 9–59. A 9.5-g bullet is fired at 760 m/s and becomes embedded in the block. How far up the incline does the block/bullet slide?(a) Ignore the friction.(b) Assume μₖ = 0.33.

Key Concepts

Conservation of Momentum

Kinetic Energy and Work-Energy Principle

Inclined Plane Dynamics

A rifle is aimed at a 2.0-kg block of wood along an inclined plane making an angle of 25°, as shown in Fig. 9–59. A 9.5-g bullet is fired at 760 m/s and becomes embedded in the block. How far up the incline does the block/bullet slide?
(a) Ignore the friction.
(b) Assume μₖ = 0.33.