Fig.1. Result of a collision.
A 750kg car collides head on with a truck 2250kg traveling opposite at the same speed.
In which direction would the wreckage move?
The answer is obvious; they will move in the direction the truck had been moving.
Reason?
Truck was several times massive than the car.
Remember it is not the size that matter but the mass.
If the car was traveling at 100km/h and the truck had a speed of 25 km/h what would happen?
Now it is not going to be obvious.
In order to determine the impact of a collision there are three factors that has to be considered.
Do you know them?
They are mass, speed and direction of movement. Instead of speed and direction, if we take velocity, this gets reduced to two; mass and velocity. This entity we refer to as MOMENTUM.
MOMENTUM = MASS X VELOCITY
P = m . v |
According to the 2nd example,
Momentum of car = 750 x 100= 75000kgms-1
Momentum of truck= 2250 x 25 = 56250 kg.ms-1
As the car has a bigger momentum the wreckage will move in the direction the car was traveling.
Elastic and non-elastic collisions.
There are two kinds of collisions.
1. Elastic collisions
In the case of an elastic collision the total kinetic energy remains without changing to heat or other forms.
2. Non-elastic collision.
In this case, part of the kinetic energy gets converted to heat , vibrations etc. In all our calculations here we shall consider them as elastic collisions.
Change in momentum = final momentum – initial momentum
Δ P = Pf - Pi |
Example 1
Fig.2. Striking a ball will change the momentum of the ball.
The figure shows a cricketer hitting a ball. The mass of the ball is 100g. (0.1kg) It reaches the bat at 20m.s
-1 and gets hit to travel in the opposite direction. If the final speed is 30 s
-1 , what is the change in momentum?
If we consider the final direction as + the initial direction is –
The initial momentum = - 0.1x 20 = 2 kg.ms-1
The final momentum is = + 0.1 x 30 = 3 ms-1
Change in momentum = +3 - ( - 2) = 5 kg.ms-1
Law of Conservation of Momentum.
In the case of an elastic collision, If two objects A and B, collide ( or explode) the sum of their original momentum is equal to the sum of their final momentum.
M ax Va final + Mb x Vb final = Ma initial x Va initial + Mb x V b initial
That is
Pa final + Pb final = Pa initial + Pb initial. |
Example 2
A trolley A of 2.0 kg moving at 1.2 ms-1 on a frictionless runway hits a still trolley B of 1.0 kg. If B gains a velocity of 1.5 ms-1
Considering this as an elastic collision, find the final velocity of A.
Let the final velocity of A be V.ms-1.
Using the formula
p a final + Pb final = p a initial Pb initial
(2. x V) + (1 x 1.5 ) = (2 x 1.2) + ( 1 x0)
2V = (2.4 - 1.5) ms-1
V = 0.9/2 ms-1
V= 0.45 ms-1
Examle 3
A white ball of 1.5 kg rolls on to a stationary grey ball of 2kg at a speed of 3ms-1. If the white one bounces back at a speed of 1 ms-1 Find the speed of the grey ball.
Working
Final Momentum = Initial momentum
(2xV) + (1.5 x-1) = (1.5x 3) + (2x0)
2V + (-1.5) = 4.5
2V = 4.5 +1.5
Therefore V = 3 ms-1
Newton’s second law states.
“ Force is directly proportional to mass of an object, while inversely proportional to acceleration”
Taking this in symbols we get :-
Force = mass x acceleration
F = ma.
As acceleration is equal to rate of change in velocity
F = mass (Vfinal - V Initial) / time
F = m ( Vf – Vi) /t
Taking time to left and removing bracket we get
Fx t = mVf – mV i
Force into time gives the change in momentum.
F.t = ΔP
|
| Impulse = Force x time = change in momentum |
What results on the object due to a force acting on it for a certain time depends on this factor.
So this is called IMPULSE. This causes the change in momentum.
1.0 There are two wagons each having a mass of 4000 kg. A is moving at a velocity of 6.ms=1 to right and B in the opposite direction at - 4 ms=1. B carries a load of 1500 kg.
Find the following:-
1.1 The momentum of A.
1.2 The momentum of B.
1.3 If they collide, the combine momentum.
1.4 The final speed of the coupled wagons.
1.5 The direction of final movement. ( to right or left)
5 x 5= 25 Marks.
2.0 A ball having a momentum of 8 kg,ms-1 strikes a wall and bounces at right angles with a momentum of 6 kg.ms-1
2.1 Find the change in momentum.
2.2 If the final velocity was 13.5 what would have been the mass of the ball.
2.3 If the ball was in contact with the wall for 0.06 seconds what would have been the magnitude and direction of the force on the wall.
5x3=15 Marks.
3.0 An explosive mix is inserted into a drilled cavity of a rock. On ignition it breaks up into 3 parts.
Part 1:- Mass= 0.1 kg. flies at a velocity of 10 ms-1 to North.
Part 2:- Mass = 0.2 kg. flies at a velocity of 10ms-1 to East.
Part 3 : …………………………… ?
Find the following:-
3.1 The total momentum just before the explosion..
3.2 The total momentum after the explosion.
3.3 The momentum of part 3 .
3.4 If the mass of part 3 is 0.1kg determine its velocity.
5x4 = 20 Marks.
4.0
A carom piece having a mass of 5g hits the board at a velocity of 4 m/s, making an angle of 45∘ with the edge. It bounces back at a speed of 3m/s.
Find the following.
i. Angular change in the direction.
ii. Original momentum.
iii. Final momentum.
iv. Change in momentum.
v. Direction of change in momentum
5x5 25 Marks.
5.0
A car having a mass of 1200 kg. speeding at 28m/s skidded and banged onto a post. The movement of the pillar and the depth of the dent caused to the car amounted to 1.3 m.
I. Tthe original momentum of the car?
Ii. The final momentum of the car?
iii.The change in momentum?
Iv. The time taken for the change?
