Edexcel iGCSE Physics (4PH1) 4.3 Work and Power Exam Style Question Paper 2B - New Syllabus

(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.