Impulse and momentum
Impulse and momentum are useful quantities when
thinking about collisions and explosions.
Impulse = Force x Time (units Ns)
Momentum = Mass x Velocity (units kgms-1)
An Impulse produces a change of momentum
When throwing
something the greater the impulse the better, in other words the biggest force possible for the
longest practicable time.
A javelin thrower will always throw the javelin starting with their
arm well behind their body. This is to give the maximum time for the force to act on the javelin
before release. A high jumper will always "sink" on the last stride before take off to allow the legs
to bend giving a longer time during take off as their legs straighten again to push them upwards.
The super gun with a long barrel will project a certain shell further than a gun with a short barrel –
the explosive gases have more time to speed it up!
Think about the impact on your feet
during running – the force is meant to be something like four times your body weight every time
your feet touch the road, hence the need for good trainers. Then there is the crash helmet – it is
designed for both protection and cushioning. It will slow down the rate of deceleration of your head
in the event of a crash!
In catching a ball you are trying to destroy its momentum and so
the time of catching is made long to reduce the forces needed to stop it. It takes the sting out of it!
A parachutist bends their knees on landing - increasing the time so reducing the force.
In
triple jumping it is vital to flex the knees as you hit the runway - rigid legs give broken bones!
Seat belts in cars are designed to give a long stopping time - much better than smashing
your head against the dashboard or the windscreen. Another place where impulse is important is
in the design of surfaces for children's playgrounds. If the children fall its important that they don't
injure themselves, in other words the surfaces they fall on must be firm enough to walk on but not
so hard that they stop in too short a time. The safe height to jump from without hurting themselves
is about 2m onto bark, sand and wet turf but onto concrete it's only 0.25 m.
Example problems
1. A girl of mass 50 kg jumps from a 1.8 m high wall and lands on the ground. Calculate the force on her if:
(a) she bends her knees and stops in 0.2 s and
(b) if she keeps her legs straight and stops in 0.05 s
(a) Velocity on impact = (2x9.8x1.8)1/2 = 5.93 ms-1
Momentum change = 50x5.93 = 297 Ns
Therefore force = F/t = 297/0.2 = 1480 N
(b) Force = 297/0.05 = 5930 N
2. A rifle of mass 3 kg fires a bullet of mass 50 gm (0.05 kg) at 100 ms-1.
(a) Calculate the recoil velocity of the rifle
(b) explain why it is better to hold the rifle into your shoulder when firing it.
(a) recoil velocity (V) = -vm/M = 100x0.05/3 = 1.67 ms-1
(b) if the rifle is in contact with your shoulder when it is fired the mass that recoils is the mass of the rifle plus your mass. If your mass is 50 kg then:
Recoil velocity = - 0.05x100/53 = 0.09 ms-1 much less then before!
3. A bullet of mass 0.05 kg moving at 100 ms-1 strikes a block of wood and is embedded in it to a depth of 5 cm. If the wood has a mass of 2.5 kg calculate:-
(a) the velocity of the block and bullet after the collision
(b) the mean retarding force of the bullet
(a) Using: velocity (V) = vm/M = 100x0.05/2.5 = 2 ms-1
(b) Using: - Kinetic energy lost = Retarding force (F) x distance
Therefore 250 -10.2 = 239.8 = Fx0.05
Retarding force = 4796 N
4. A car of mass 1000 kg travelling from west to east along a road at 15 ms-1 collides with a van of mass 3000 kg travelling in the opposite direction. If they both come to rest after the collision what was the initial velocity of the van?
Momentum after the collision = 0 so momentum before collision must also equal 0.
Therefore :
Momentum before collision = momentum of car + momentum of van = 1000x15 + 3000v where v is the velocity of the van.
This gives 1000x15 + 3000v = 0 and so v = - 1000x15/3000 = -5 ms-1 the negative sign meaning that it was travelling in the opposite direction i.e. east to west.
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