A pellet gun fires a small lead pellet of mass 0.0010 kg horizontally into a block of modelling clay of mass 0.25 kg resting on a shelf 1.2 m above the floo...

(a) Time to fall from shelf to floor [2]

The clay-pellet combination is launched horizontally, so the vertical motion is free fall from rest. Using \(s = \tfrac{1}{2}gt^2\):

\(1.2 = \tfrac{1}{2} \times 10 \times t^2\)

\(t^2 = \dfrac{1.2}{5.0} = 0.24\)

\(t = \sqrt{0.24} = 0.49 \text{ s}\) [2]

(b) Horizontal speed leaving the shelf [2]

Horizontally, there is no acceleration (no air resistance), so the horizontal speed is constant throughout the fall:

\(v_{\text{horizontal}} = \dfrac{\text{horizontal distance}}{\text{time}} = \dfrac{0.80}{0.49} = 1.63 \text{ m/s}\) (accept 1.6 m/s) [2]

This is the speed of the clay-pellet combination immediately after the pellet embeds in the clay.

(c) Speed of the pellet before impact [3]

The clay was initially stationary, and the pellet embeds in it (perfectly inelastic collision). By conservation of momentum:

\(m_{\text{pellet}} \times v_{\text{pellet}} = (m_{\text{pellet}} + m_{\text{clay}}) \times v_{\text{after}}\)

\(0.0010 \times v = (0.0010 + 0.25) \times 1.63\) [1]

\(0.0010 \, v = 0.251 \times 1.63 = 0.409\) [1]

\(v = \dfrac{0.409}{0.0010} = 409 \text{ m/s}\) (accept 410 m/s) [1]

The projectile analysis (parts a and b) gives the speed after the collision, and conservation of momentum then works backward to find the pellet's original speed. This two-stage technique is commonly used in forensic ballistics.