A pellet with velocity 200 m s⁻¹ and mass 5.0 g is fired vertically upwards into a stationary block of mass 95.0 g - AQA - A-Level Physics - Question 22 - 2020 - Paper 1

The total mass of the system after the pellet embeds into the block is:

$m_{total} = m_{pellet} + m_{block} = 0.005 \, kg + 0.095 \, kg = 0.1 \, kg$.

By applying the principle of conservation of momentum, the velocity of the block and pellet combined after the collision is given by:

$mv_{initial} = m_{total}v_{final}$

Substituting the known values:

$1 \, kg \, m/s = 0.1 \, kg \cdot v_{final}$.

This simplifies to:

$v_{final} = \frac{1 \, kg \, m/s}{0.1 \, kg} = 10 \, m/s$.

To find the maximum vertical displacement, we can use the formula for maximum height reached under uniform acceleration due to gravity:

$h = \frac{v_{final}^2}{2g}$ Where $g = 9.81 \, m/s^2$ is the acceleration due to gravity.

Substituting in the values:

$h = \frac{(10 \, m/s)^2}{2 \cdot 9.81 \, m/s^2} \approx 5.1 \, m$.