# Evan's Space

## PP N2010 P1 Q4

Solutions: Option C

When a body is released from rest, the only force acting on the body is its weight due to gravity. Both bodies experience acceleration due to gravity. Hence for both bodies (regardless of mass), will have the same initial acceleration of 10 m/s2.

As the two bodies are of the same size and shape, they will experience the same air resistance for any particular speed. As speed increases, air resistance increases.

For terminal velocity to be reached, air resistance has to be equal to the weight. Since weight is greater for the ball with larger mass, the air resistance has to be bigger. Thus, the ball has to accelerate more (air resistance increases with speed) for the larger air resistance to be equal to the weight. Hence the ball with larger mass will have larger terminal velocity.

[NOTE]
Do not confuse ‘speed of the body is independent of the mass’ as learned in Work Done, Energy and Power. This concept is based on the assumption that there is no air resistance. So not applicable in this question as for terminal velocity to occur, air resistance must be present.

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## Kinematics Question – Using gradient and area under graph

Solutions: Option B

## Object falling at freefall

A particle released from rest at O falls freely under gravity passes through P and Q as shown.

If the particle takes 4 s to move from P to Q where PQ = 120 m, how long does it take to fall from O to P?

A) 1s     B) 2s     C) 4s     D) 8s

Solutions: Option A

## Rocket with forward thrust lasting 1.5 s

An experiment rocket model of mass 0.05 kg was launched from the ground, vertically into the air with a propelling force of 4.5 N. The air resistance is assumed to be negligible. The rocket carries propellant only enough for 1.5 s of the upward flight. The rocket then crashed to the ground after some time. (Take g = 10 ms-2)

a) Calculate the weight of the rocket model.   Ans: 0.5 N

b) Calculate the acceleration of the rocket, which is assumed to be uniform, during the first 1.5 s of flight.    Ans: 80 ms-2

c) What is the maximum velocit of the rocket?  Ans: 120 m/s

d) What is the upward acceleration of the rocket after 1.5 s of flight?   Ans: – 10 ms-2

e)Draw a clearly labeled velocity-time and displacement-time graph for the whole flight of the rocket from its take-off to its crashing to the ground.

## Car A and Car B – Find the time one overtakes the other

The speed-time graph of Car A and Car B, along a straight road over 4 seconds is shown below.

a) Calculate the acceleration of Car A and Car B over the 4 seconds.
Ans: a of Car A = 3 ms-2 and a of Car B =0.75 ms-2

Car A overtakes Car B at time t seconds.

b) Derive two separate expressions for the velocities of Car A and Car B at the point when Car A overtakes Car B, in terms of t.
Ans: Va = 3t    and    Vb = 3 + 0.75t

c) Calculate the time t when Car A overtakes Car B.
Ans: t = 2.67s

Solutions for (a) and (b)

Solutions for (c)

## Vertical Resultant Force of ball rolling off platform

A ball of mass 5 kg rolls along a smooth horizontal surface until it falls off the edge at time = 3s and touches the lower surface at t = 5s. Sketch the graph which represents how the resultant vertical force F acting on the ball varies with time as the ball moves from A to B.

The question only ask for vertical resultant force, hence there is no need to consider the horizontal forces. Anyway the ball is moving horizontally at constant speed, hence no acceleration, forces are balanced. It will move at constant speed (Newton’s 1st law) When the ball is rolling on the upper/lower platform, the weight is equal to the normal force acting on the ball by the platform, hence net force is 0N.