Physics

1. 1.

   A 0.200 kg ball moves east at 12.0 m s^-1 and rebounds west at 8.00 m s^-1 in 0.0400 s. Calculate the impulse on it and mean force, with directions.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Reversal changes the sign of velocity; the speed difference alone is not the momentum change.
   2. Divide the signed impulse -4.00 N s by the contact time to obtain -100 N.

   Marking points

   • Taking east positive, delta p = 0.200(-8.00 - 12.0).
   • Impulse is 4.00 N s west.
   • Mean force is 100 N west (3 significant figures).

   Examiner tip: Use a signed velocity for the rebound before subtracting.

2. 2.

   A 0.50 kg mass travels uniformly around a horizontal circle of radius 0.80 m at 4.0 m s^-1. Calculate the resultant force and explain why acceleration is nonzero.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Centripetal describes the direction of the resultant, not an extra force added to the others.
   2. Multiply the radial acceleration by mass; no tangential acceleration is required for uniform speed.

   Marking points

   • a = v^2/r = 20 m s^-2 towards the centre.
   • Resultant force is 10 N towards the centre (2 significant figures).
   • Velocity changes direction even though speed is constant.

   Examiner tip: Constant speed is not constant velocity on a curved path.

3. 3.

   On a frictionless track a 0.60 kg cart at 5.0 m s^-1 hits a stationary 0.40 kg cart and they stick. Calculate their velocity and kinetic energy lost. External impulse during collision is negligible.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Sticking makes this an inelastic collision: total momentum is conserved but kinetic energy is not.
   2. Compute kinetic energies separately using the moving mass before and combined mass after.

   Marking points

   • Momentum conservation: 0.60*5.0 = (0.60 + 0.40)v.
   • v = 3.0 m s^-1 in the original direction.
   • Initial KE = 7.5 J; final KE = 4.5 J; loss = 3.0 J (2 significant figures).

   Examiner tip: Do not impose kinetic energy conservation for sticking carts.

4. 4.

   A road is banked at 20.0 degrees around a curve of radius 50.0 m. Find the speed requiring no friction. Use g = 9.81 m s^-2 and model the vehicle as a particle.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Resolve the normal reaction, which is tilted towards the centre because the road is banked.
   2. Mass cancels when the two component equations are divided; use degree mode for 20.0 degrees.

   Marking points

   • Vertical balance: N cos theta = mg; horizontal: N sin theta = mv^2/r.
   • Divide to obtain v^2 = rg tan theta.
   • v = 13.4 m s^-1 (3 significant figures).

   Examiner tip: The centripetal resultant is horizontal, while vertical forces balance.

5. 5.

   A bead is released from rest at height h above the bottom of a smooth vertical circular track of radius R = 0.500 m. Find the minimum h to maintain contact throughout the inside loop and the bottom speed at that h. Use g = 9.81 m s^-2; ignore bead size and losses.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. At the top gravity and the normal act inward; contact fails if a negative normal would be required.
   2. Use the smallest permissible top speed in the energy equation, then transfer all initial potential energy into kinetic energy at the bottom.

   Marking points

   • At the top, marginal contact means N = 0 and v_top^2 = gR.
   • mgh = mg(2R) + mgR/2, so h = 5R/2 = 1.25 m.
   • v_bottom = sqrt(5gR) = 4.95 m s^-1 (3 significant figures).

   Examiner tip: Reaching the top with zero speed is insufficient for contact on an inside loop.

6. 6.

   A stationary object splits into fragments of masses 2.0 kg and 3.0 kg on a horizontal frictionless surface. Exactly 30 J becomes kinetic energy; external impulse is negligible. Calculate each speed and describe their directions, to 2 significant figures.

   [3 marks]

   Answer explanation

   Draft walkthroughs are based on marking guidance, not independently verified derivations.

   1. Substitute v2 = 2v1/3 into the energy equation, giving (5/3)v1^2 = 30 and v1^2 = 18.
   2. The lighter fragment is faster; equal momentum does not imply equal kinetic energy.

   Marking points

   • Equal and opposite momenta give 2.0v1 = 3.0v2.
   • Use (1/2)*2.0v1^2 + (1/2)*3.0v2^2 = 30.
   • Speeds are 4.2 m s^-1 for 2.0 kg and 2.8 m s^-1 for 3.0 kg, in opposite directions.

   Examiner tip: Conserve momentum, not kinetic energy; internal stored energy supplies the 30 J.

Original, unofficial practice questions from SubjectScout — not official past-paper questions.