A ball with mass \(m\) strikes a wall with speed \(v\). Assume that the collision is elastic. If the ball bounces back with the same speed \(v\), the magnitude of the change in momentum will be ...
About this question
Question 1.4 of Technical Sciences Grade 12 Paper 1, NSC Preparatory Examination (Eastern Cape DoE, September 2025). It tests the Term 1 ATP topic Momentum & Impulse Momentum.
- TermTerm 1
- ATP topicMomentum & Impulse Momentum
- PaperPaper 1
- Marks2 marks for the correct letter
A: \(2mv\) in the opposite direction.
\(\Delta p = m(-v) - mv = -2mv\), so the change is \(2mv\) in the direction opposite to the original motion.
More Questions On Momentum & Impulse Momentum
Compare the total kinetic energy before and after the collision.
\[ \sum E_{ki} = \tfrac{1}{2}(1500)(33{,}33)^2 + \tfrac{1}{2}(1100)(16{,}67)^2 \approx 986\,006\ \text{J} \]\[ \sum E_{kf} = \tfrac{1}{2}(1500)(20{,}3)^2 + \tfrac{1}{2}(1100)(5{,}32)^2 \approx 324\,634\ \text{J} \]\( \sum E_{ki} \neq \sum E_{kf} \), so kinetic energy is not conserved.
The collision is inelastic.
C: in an isolated system the total linear momentum before the collision is equal to the total linear momentum
In an isolated system the total momentum before a collision equals the total momentum after it.
Take the direction towards the tree as negative for the force (it opposes the motion).
\[ F_{net}\Delta t = \Delta p = m(v_f - v_i) \]\[ -57\,500\,\Delta t = 1150(0 - 15) \]\( \Delta t = 0{,}30\ \text{s} \)
\( p = 18\,337\ \text{kg}\cdot\text{m}\cdot\text{s}^{-1} \), west
Three points explain how an airbag reduces injuries:
- The impulse (change in momentum) of the occupant stays the same.
- The airbag increases the contact time during the crash.
- A longer contact time gives a smaller force on the occupant, so the injuries are less severe.