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 momentum before and after the collision should remain the same. Let $M$ be the mass of the block (95.0 g = 0.095 kg) and $V$ be the velocity of the block after the collision:

$1 ext{ kg m s}^{-1} = (m + M)V$

Substituting the values, $1 = (0.005 + 0.095)V$

Which simplifies to: $1 = 0.1V$

This gives: $V = 10 ext{ m s}^{-1}$.

At the maximum height, all kinetic energy is converted to potential energy. The kinetic energy (KE) can be calculated using:

$KE = \frac{1}{2}mv^2$

The potential energy (PE) at height $h$ is given by:

$PE = mgh$

Using $m = 0.1$ kg (combined mass), $v = 10$ m s⁻¹, and $g = 9.81$ m s⁻², we equate KE and PE:

$\frac{1}{2}(0.1)(10)^2 = 0.1(9.81)h$

Thus: $5 = 0.981h$

Solving for $h$ gives: $h = \frac{5}{0.981} \approx 5.1 ext{ m}$.