Edexcel iGCSE Physics (4PH1) 1.4 Momentum Exam Style Question Paper 1B - New Syllabus
Question
The diagram shows the collision between two balls, A and B. The masses and velocities of both balls are shown before and after the collision. Ball B is stationary before the collision.
(a) When the balls collide, ball B applies a force on ball A, which causes the velocity of ball A to change. Ball A also applies a force on ball B during the collision. Describe how the force applied on ball A compares with the force applied on ball B during the collision.
(b) Calculate the momentum of ball A before the collision.
(c) Show that the velocity, \(v\), of ball B after the collision is about \(0.6\,\mathrm{m\,s^{-1}}\).
(d) A collision is considered elastic if the total kinetic energy before the collision is equal to the total kinetic energy after the collision. Using data from the diagram, deduce whether this collision is elastic.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.28P: Force, Momentum, and Time — part (a)
• 1.29P: Newton’s Third Law — part (a)
• 1.13–1.15: Scalars, Vectors, and Force as a Vector Quantity; Resultant Forces — part (a)
• 4.14: Kinetic Energy — part (d)
• 4.15: Conservation of Energy in Mechanical Systems — part (d)
▶️ Answer/Explanation
(a) Forces during the collision [2 marks]
- The force applied by ball A on ball B is equal in magnitude to the force applied by ball B on ball A.
- The two forces act in opposite directions.
This is an example of Newton’s Third Law. The forces form an action-reaction pair and act on different balls.
(b) Momentum of ball A before the collision [2 marks]
Use the momentum equation:
\(p=m v\)
From the diagram:
\(m=0.018\,\mathrm{kg}\)
\(v=4.9\,\mathrm{m\,s^{-1}}\)
Therefore:
\(p=0.018\times4.9\)
\(p=0.0882\,\mathrm{kg\,m\,s^{-1}}\)
So the momentum of ball A is approximately:
\(\boxed{0.088\,\mathrm{kg\,m\,s^{-1}}}\)
(c) Velocity of ball B after the collision [4 marks]
Momentum is conserved in the collision.
Total momentum before = total momentum after
Ball B is initially stationary, so its initial momentum is zero.
Therefore:
\(0.088=0.018\times(-3.5)+0.265v\)
Rearranging:
\(0.088+0.063=0.265v\)
\(v=\dfrac{0.151}{0.265}\)
\(v\approx0.57\,\mathrm{m\,s^{-1}}\)
Therefore:
\(\boxed{v\approx0.6\,\mathrm{m\,s^{-1}}}\)
(d) Determining whether the collision is elastic [4 marks]
For an elastic collision, the total kinetic energy before the collision must equal the total kinetic energy after the collision.
Use:
\(E_{\mathrm{k}}=\dfrac{1}{2}mv^2\)
Kinetic energy before the collision:
Ball B is stationary, so its initial kinetic energy is zero.
\(E_{\mathrm{k,before}}=\dfrac{1}{2}(0.018)(4.9)^2\)
\(E_{\mathrm{k,before}}\approx0.216\,\mathrm{J}\)
Kinetic energy after the collision:
For ball A:
\(E_{\mathrm{k,A}}=\dfrac{1}{2}(0.018)(3.5)^2\approx0.110\,\mathrm{J}\)
For ball B:
\(E_{\mathrm{k,B}}=\dfrac{1}{2}(0.265)(0.57)^2\approx0.043\,\mathrm{J}\)
Therefore:
\(E_{\mathrm{k,after}}\approx0.110+0.043=0.153\,\mathrm{J}\)
Since \(0.216\,\mathrm{J}\neq0.153\,\mathrm{J}\), the total kinetic energy is not conserved.
Therefore, the collision is not elastic. It is an inelastic collision.
Question
The photograph shows a dummy during a test of the safety features of a car in a collision.
Before the collision, the dummy in the car is travelling at a velocity of \(14\,\mathrm{m\,s^{-1}}\). The dummy has a momentum of \(1100\,\mathrm{kg\,m\,s^{-1}}\).
(a)(i) State the formula linking momentum, mass and velocity.
(ii) Show that the mass of the dummy is approximately \(80\,\mathrm{kg}\).
(b) The dummy is brought to rest during the collision by a mean force of \(15\,\mathrm{kN}\). Calculate the time taken for the dummy to be brought to rest in the collision.
(c) The car being tested is fitted with airbags. Using ideas about momentum, explain how an airbag reduces the force experienced by the dummy in the collision.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.28P: Force, Momentum, and Time — parts (b) and (c)
▶️ Answer/Explanation
(a)(i) Momentum formula [1 mark]
The formula linking momentum, mass and velocity is:
\(p=mv\)
(a)(ii) Mass of the dummy [2 marks]
Using \(p=mv\), rearrange to make mass the subject:
\(m=\dfrac{p}{v}\)
Substitute the values:
\(m=\dfrac{1100}{14}\)
\(m=78.6\,\mathrm{kg}\)
Therefore, to an appropriate number of significant figures:
\(\boxed{m\approx79\,\mathrm{kg}}\)
This is approximately \(80\,\mathrm{kg}\), as required.
(b) Time taken to stop [3 marks]
Use the relationship between force, change in momentum and time:
\(F=\dfrac{\Delta p}{t}\)
The dummy is brought to rest, so its final momentum is zero. Therefore, the magnitude of the change in momentum is \(1100\,\mathrm{kg\,m\,s^{-1}}\).
Convert the force into newtons:
\(15\,\mathrm{kN}=15000\,\mathrm{N}\)
Rearrange:
\(t=\dfrac{\Delta p}{F}\)
Substitute:
\(t=\dfrac{1100}{15000}\)
\(t=0.0733\,\mathrm{s}\)
Time taken: \(\boxed{0.073\,\mathrm{s}}\)
(c) How an airbag reduces the force [2 marks]
- The airbag increases the time over which the dummy is brought to rest.
- For the same change in momentum, increasing the collision time reduces the rate of change of momentum, and therefore reduces the force experienced by the dummy.
This follows from \(F=\dfrac{\Delta p}{\Delta t}\). For the same \(\Delta p\), a larger \(\Delta t\) gives a smaller force.