Newton's Laws Revision Set 2

Newton's Laws Revision Set 2

9th - 10th Grade

10 Qs

quiz-placeholder

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Newton's Laws Revision Set 2

Newton's Laws Revision Set 2

Assessment

Quiz

Physics

9th - 10th Grade

Practice Problem

Hard

Created by

S Doran

Used 2+ times

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10 questions

Show all answers

1.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A parachutist is falling through the air at terminal velocity. Which statement about the parachutist is correct?

Every force acting on the parachutist is equal to zero and his acceleration is equal to zero.

Every force acting on the parachutist is equal to zero and his velocity is equal to zero.

The resultant force acting on the parachutist is equal to zero and his acceleration is equal to zero.

The resultant force acting on the parachutist is equal to zero and his velocity is equal to zero.

Answer explanation

At terminal velocity, drag = weight, so zero resultant force so zero acceleration

2.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A stone of mass 0.12 kg is fired from a catapult. The velocity of the stone changes from 0 to 5.0 m / s in 0.60 s.

What is the average resultant force acting on the stone while it is being fired?

1.0 N

2.5 N

3.6 N

8.3 N

Answer explanation

acceleration = change in velocity ÷ time

5 m/s ÷ 0.6 s

Force = mass x acceleration

= 0.12 kg x 5 m/s ÷ 0.6 s

= 1 N

3.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A ball of mass 0.16 kg is moving forward a velocity of 0.50 m/s. A second ball of mass 0.10 kg is stationary. The first ball strikes the second ball. The second ball moves forwards at a speed of 0.50 m/s. What is the velocity of the first ball after the collision? (Hint: conservation of momentum)

0.0 m/s

0.19 m/s

0.31 m/s

0.50 m/s

Answer explanation

Momentum before collision

= 0.16 kg x 0.5 m/s = 0.08 kgm/s

Momentum after collision = 0.08 kgm/s

(0.16 kg x ? m/s) + (0.1 kg x 0.05 m/s) =

(0.16 kg x ? m/s) + 0.05 kgm/s

0.16 kg x ? m/s = 0.03 kgm/s

velocity = momentum ÷ mass

= 0.03 kgm/s ÷ 0.16 kg

= 0.1875 m/s

4.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A student cycles along a level road at a speed of 5.0 m/s. The total mass of the student and bicycle is 120 kg. The student applies the brakes and stops. The braking distance is 10 m. What is the average braking force? (Hint work out acceleration from v & s)

150 N

300 N

15 000 N

30 000 N

Answer explanation

(final velocity)2-(initial velocity)2=

2x acceleration x distance

Rearranging to make a the subject:

a = (v2- u2)/2s v = 0m/s, s = 10 m

a = -25/20 = 1.25 m/s2

force = mass x acceleration

= 120 kg x 1.25 m/s2 = 150 N

5.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

What is the unit of power?

joule

newton

pascal

watt

Answer explanation

watt, (W) = J/s

6.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

A box of mass 2.0 kg is pulled across the floor by a force of 6.0 N. The frictional force acting on the box is 1.0 N.

What is the acceleration of the box?

0.40 m/s2

2.5 m/s2

3.0 m/s2

3.5 m/s2

Answer explanation

Resultant force on box = 5 N (6-1)

acceleration = force ÷ mass

7.

MULTIPLE CHOICE QUESTION

3 mins • 1 pt

Which moving body has a resultant force acting on it?

a diver rising vertically through water at constant speed

an aircraft circling an airport at constant speed

a train going up a straight incline at constant speed

a parachutist descending vertically at terminal velocity

Answer explanation

A resultant force is needed to change velocity (SPEED and/or DIRECTION)

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