IGCSE physics force and motion worksheet with numerical problems | with solution
Last updated on April 17th, 2021 at 09:36 am
Here you will get 2 sets of IGCSE physics force and motion worksheet with numerical problems with solution. You will get ample scope to test your understanding of motion equations and force concepts.
The solutions of worksheet set 1 are given on this page itself. And the solution of the set 2 numerical worksheet is linked with the individual problem of set 2.
IGCSE physics force and motion worksheet | IGCSE physics numerical-set 1
1) A force of 1000 Newton is applied on a 25 kg mass for 5 seconds. What would be its velocity?
2) A force is applied to a mass of 16 kg for 3 seconds. As the force is removed the mass moves 81 meters in 3 seconds. What was the value of the force applied?
3) A mass of 50 kg was moving with a velocity of 400 m/s. A force of 40000 N is applied to the mass and its velocity is reduced to 50 m/s after some time. What is the distance traveled by the mass during this period?
4) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 second. What was the force applied to the ball when it hits the wall?
5) Momentum of an object changes from 100 kg m/s to 200 kg m/s in 2 seconds. What is the force applied to it?
6) A force of 200 N is applied on a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?
7) A mass of 1 kg is moving from east to west with a velocity of 10 m/s. A force is applied to it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?
solutions for set 1 – IGCSE physics worksheet solved
1) Force of 1000 Newton is applied on a 25 kg mass for 5 seconds. What would be its velocity?
Solution:
F =1000 N
m=25kg
t= 5 sec
u=0
v=?
Acceleration = a = F/m = 1000/25 = 40 m/s2
v = u + at =0 + 40X5 = 200 m/s
2) A force is applied on a mass of 16 kg for 3 seconds. As the force is removed the mass moves 81 meter in 3 seconds. What was the value of the force applied?
Solution:
first 3 secs (Force is present) –> next 3 secs (F is not there, so uniform velocity)
In the next 3 seconds, the mass travels with uniform velocity (as there is no force that means no acceleration).
So the velocity in the next 3 secs:
V = 81/3 = 27 m/s
This velocity was attained in the first 3 seconds when the force was present.
So in the first 3 seconds, velocity changes from 0 to 27m/s.
Therefore the acceleration in first 3 seconds caused by the force present = (27-0)/3 = 9 m/s².
So the force applied on the mass for first 3 secs= mass x acc = 16 x 9 = 144 N
3) A mass of 50 kg was moving with a velocity 400 m/s. A force of 40000 N is applied on the mass and its velocity is reduced to 50 m/s after some time. What is the distance travelled by the mass during this period?
Solution:
Mass = 50 kg
u = 400 m/s
v= 50 m/s
F = 40000 N
t = unknown
Distance travelled in time t =?
Acceleration = a = F/m = 40000/50 = 800 m/s2
Again, acceleration a = change of velocity /t
or, t = change of velocity/a = 350/800 sec
Distance travelled: s = ut – (1/2)a t2
= 400 * (350/800) – (1/2) 800. (350/800)2
= 175 – 76.5 = 98.5 m
Physics solutions: continued
4) A ball of 150 g mass moves with 12 m/s and bounces back after hitting a wall with 20 m/s after a small duration of 0.01 second. What was the force applied on the ball when it hits the wall?
Solution:
Physics problems for class 9 with solutions – physics problems solved
Mass = 150 gram = 150/1000 kg = 0.15 kg
Change of velocity = 20 – (-12) = 32 m/s
T = 0.01 sec
Acceleration = change in velocity / time = 32/0.01 = 3200 m/s2
Force = mass X acc = 0.15 X 3200 N= 480 N.
solutions: continued
5) Momentum of an object changes from 100 kgm/s to 200 kgm/s in 2 seconds. What is the force applied on it?
Solution:
Note:Force = ma = m(v-u)/t = (mv – mu)/t = change of momentum /time = rate of change of momentum
Force= change of momentum /time = (200-100)/2 = 100/2 = 50 N
6) A force of 200 N is applied on a body and its velocity changes from 5m/s to 10m/s in a second. What is the mass of the body?
Solution:
mass = force/acceleration
here acceleration =a = (10-5)/1 = 5 m/s2
so, mass = 200/5 kg = 40 kg
solutions: continued
7) A mass of 1 kg is moving from east to west with a velocity 10 m/s. A force is applied on it for 2 seconds and its velocity becomes 5 m/s. What is the value and direction of the force?
Solution:
Mass = 1kg
velocity = 10 m/s (east to west)
Because of the force the velocity reduces to 5 m/s in 2 secs.
acceleration = change of vel/time=5/2 = 2.5 m/s2
Force = mass X acceleration= 1 X 2.5 = 2.5 N.
As the velocity reduces because of force, (negative acc i.e. retardation) hence the direction of the force would be opposite to the initial direction. So the direction of the force will be west to east.
Physics Worksheet 2 | Force & motion Assignments IGCSE – Set 2
11) A force acts for 10 s on a stationary body of mass 100 kg after which the force ceases to act. The body moves through a distance of 100 m in the next 5 s. Calculate (i) the velocity acquired by the body, (ii) the acceleration produced by the force, and (iii) the magnitude of the force.
[ Solution: click this link for solution Q11 ]
12) A force acts for 0.1 sec on a body of mass 2 kg initially at rest. The force is then withdrawn and the body moves with a velocity of 2 m/s. Find the magnitude of the force.
Solution: click this link for solution Q12
13) A cricket ball of mass 100 g moving at a speed of 30 m/s is brought to rest by a player in 0.03 s. find the average force applied by the player.
Solution: click this link for solution Q13
14) A body of mass 500 g, initially at rest, is acted upon by a force that causes it to move a distance of 4 meters in 2 sec. Calculate the force applied.
Solution: click this link for solution Q14
15) A force causes an acceleration of 10 m/s^2 in a body of mass 500 g. What acceleration would be caused by the same force in a body of mass 5 kg?
Solution: click this link for solution Q15
16) A bullet of mass 50 g moving with an initial velocity of 100 m/s, strikes a wooden block and comes to rest after penetrating a distance of 2 cm in it.
Calculate:
(i) Initial momentum of the bullet (ii)final momentum of the bullet (iii) retardation caused by the wooden block, and (iv) resistive force exerted by the wooden block
Solution: click this link for solution Q16