Year 12 Revision Test 3

Year 12 Revision 5.Name:

There are 8 questions, each is worth one mark unless otherwise stated. Time = 16 minutes.

A boy tries to move a heavy box across the floor. He applies an ever increasing force, but is unable to move it. Figure 5 below illustrates the situation.

Question 1

What is the work done by the force Fb as the boy unsuccessfully tries to pull the box?

Question 2

Which one of the graphs( A.-F.) below best shows the variation of the force of friction on the box (Ff) as the force applied by the boy (Fb) varies?

A skydiver drops from a plane. The graph in Figure 6 below shows the force of air resistance on the skydiver as a function of the distance fallen. After falling a distance of 500 m, the skydiver has reached terminal velocity and continues to travel at constant speed.

Question 3

What is the value of the net force on the skydiver after falling 500 m?

Question 4

Calculate the mass of the skydiver.

Question 5

Calculate the acceleration of the skydiver at the instant when the distance fallen is 100m.

Question 6

Estimate the magnitude ofthe work done by the air resistance force on the skydiver while falling the first 100m.

A ball of mass 0.100kg is dropped vertically on to a hard surface reaching a speed of 12ms-1 just before hitting the surface. It rebounds vertically at an initial speed of 8.0 m s-1. The contact between ball and surface lasts for a time of 0.0050 s.

Question 7

Calculate the magnitude of the change in momentum of the ball during the time of contact.

Question 8

Calculate the magnitude of the average net force exerted on the ball during the collision.

Question 1 solution

Work done is F × d, but here the distance moved is zero. So the work done is zero.

Question 2 solution

The force that the boy exerts is ever increasing, but the box doesn't actually move. This means that the friction force must be equal and opposite to the force applied by the boy. Graph E shows that when the boy is exerting a force of +60N, the friction force is -60N.

Question 3 solution

After falling 500 m the skydiver is travelling a terminal velocity. This means that they are not accelerating. Therefore the sum of the forces on them must be zero.

Question4 solution

92kg

At 500 m the weight force must be the magnitude of the resistance, but in the opposite direction.

 W = mg = 900N m = 900  9.8 = 91.8 kg

Question 5 solution

3.3 m/s2

When they skydiver has fallen 100m, the resistive forces = 600N. The weight force down = 900N.

 the net force down = 900 - 600 = 300N.

Using F = ma300 = 91.8 × a

 a = 300  91.8

 a = 3.27 m/s2

Question 6 solution

4.0 x 104J

The work done is the area under the force v displacement graph. It is best to count squares.

Each square = 100 × 50 = 5 000J.

For the first 100 m, there are ~ 8 squares. WD = 5 000 × 8 = 40 000J

Question 7 solution

2.0 kgm/s

The change in momentum is given by pf -pi = mv. v = final - initial.

v = 8 m/s

u = 12 m/s -u

v - u = 20 m/s up.

p = 0.100 × 20

= 2.0 kgm/s

Question 8 solution

4.0 × 102 N

Use F t = m v F = = 400 N