This is a suggested solution to the GCE O Level Pure Physics examination in Year 2014, Paper 2, Section B, Question 10.

## Suggested Solution

(a) For the bag to accelerate, there must be a resultant force that is greater than zero.

The resultant force is a result of the force P and the opposing frictional force of 0.5N, that is:

Resultant Force, F = P-0.5N

Hence, P must be greater than 0.5N for the resultant force to be greater than zero.

(b) (i)The equation for the resultant force, F is:

F=ma

(b) (ii)Selecting the point where P=2.5N, the acceleration, a=1.0 m/s^{2}

Resultant Force, F =2.5N -0.5N =2.0N

Since, F=ma

Then, mass, m = 2.0N/(1.0 m/s^{2})

=2.0 kg

(c) (i) At t=3.0s, a=1.25 m/s^{2}

Therefore, 1.25 m/s^{2} =[ (velocity at t=3.0s) – 0)]/3.0

Hence, velocity at time t=3.0s is 3.75 m/s

(c) (ii) When P is reduced to 0.5N, the acceleration is zero and there is no more resultant force to cause the bag to accelerate. However, according to Newton’s First Law of Motion, the bag will continue its motion at constant velocity of 3.75m/s.

(d) The force is the reactive force that acts on the student by the bag. Its magnitude is equal to P but in opposite direction of P.