GCSE level Physics exam revision notes on Forces & Motion Part 6

Forces and Newton's Laws of Motion Part 6.6

More advanced calculations involving momentum and Newton's 2nd and 3rd Laws of motion, including the concept of impulse

you need to be able to use F = ma and impulse = F x ∆t = ∆mv

Q6.1 What explosive force is required to accelerate a 20 g bullet in a gun from 0 m/s to 300 m/s in 0.15 seconds.

20 g = 20/1000 = 0.02 kg, u = 0 m/s, v -= 300 m/s, ∆t = 0.15 s

F = ∆mv / ∆t = (mv - mu) / ∆t

F = {(0.02 x 300) - (0.02 x 0)} / 0.15

F = (0.02 x 300) / 0.15 = 6.0 / 0.15 = 40 N

It is important that you can solve these sorts of problems in different ways, depending on what information you get and how you are expected to deduce things from it e.g.

Using the same information you can solve this problem using F = ma directly.

a = 300 / 0.15 = 2000 m/s², m = 0.02 kg, F = ma, F = 0.02 x 2000 = 40 N

This is the approach I took on my stopping distances, impact forces - example calculations

Q6.2 What force must a tennis player generate with a tennis racket exert on a 60 g tennis ball to accelerate it from rest to a speed of 50 m/s in 10 milliseconds (10 ms).

mass of tennis ball = 60 g = 60/1000 = 0.06 kg, time force applied = 10 ms = 10/1000 = 0.01 s

∆p = ∆mv = F x ∆t = impulse  (this comes from Newton's 2nd Law equation F = m∆v/∆t)

mv changes from zero to 0.06 x 50 = 3.0 kg m/s

therefore: 3.0 = F x 0.01,  F = 3.0 / 0.01 = 300 N

 

Q6.3 An object experiences a change in momentum of 30 kg m/s.

If the change takes place in 5 seconds, what force on the object produced this change in momentum?

F = ma = m∆/∆t = ∆mv / ∆t = 30 / 5 = 6 N

Q6.4 A force of 500 N is applied to an object to change its momentum by 2000 kg m/s.

For how long was the force applied?

F = ma = m∆v/∆t = ∆mv/∆t

∆t = ∆mv / F = 2000 / 500 = 4 seconds

Q6.5 A 0.80 g lead pellet fired from an airgun experiences a force of 4.8 N in a time interval of 25 milliseconds - the time it takes for the pellet to travel down the barrel.

(a) Calculate the speed at which the pellet leaves the gun.

F = ma = m∆v / ∆t = m(v-u) / ∆t

m = 0.8 / 1000 = 0.0008 kg,  ∆t = 25 / 1000 = 0.025 s (25 milliseconds), v = ?, u = 0 m/s

F = mv / ∆t

 v = ∆t x F / m = 0.025 x 4.8  / 0.0008 = 150 m/s

(b) If the airgun recoils at a speed of 0.05 m/s, what is the mass of the airgun?

You can solve this part of the problem two ways

(i) From the law of conservation of momentum

momentum of airgun = momentum of pellet

m x 0.05 = 0.0008 x 150

m = 0.0008 x 150 / 0.05 = 2.4 kg

(ii) From Newton's 3rd Law, which tells you that the force on the gun is equal in size to the force on the lead pellet.

F = ma, m = F / a

m =? kg, F = 4.8 N, a = ∆v / ∆t = 0.05 / 0.025 = 2.0 m/s

m = 4.8 / 2.0 = 2.4 kg

For other questions See calculation parts in section 5 including F = ma

INDEX physics notes: Conservation of momentum, elastic and non-elastic collisions, Newton's 2nd law calculations - problem solving