Question

An object of mass 1.5kg travelling in a straight line with a velocity of 5m/s collides with a wooden block of mass 5kg resting on the floor. This object sticks with the wooden block after collision and both move together in a straight line. Calculate:
(a)the total momentum before collision.
(b) total momentum after collision
(c) the velocity of the combination of these objects after collision

Answer 1

When a lower mass object strikes a larger mass object it will move in the same direction but it will have lower momentum due to decreased velocity. because the object when strikes the greater mass object it loses some of energy in overcoming the resistence offered by the greater mass object kept in its path.
Now see,
the momentum of the smaller object is 1.5×5=7.5 kg m/s
And the kinetic energy possesed by it will be;
K=p²/2m
⇒K=56.25/3=18.75 Joules.
Energy required to move the greater mass object =its mass = 5 Joules.
So the energy waisted by the smaller mass = 5x as the greater mass will occupy some speed x also, So, the lost energy by the smaller will be=5x Joules
The energy left with the smaller mass= (18.75-5x) Joules.
And so the momentum of the smaller mass we will calculate by using the formula p=√2mK
⇒√(2×1.5×(18.75-5x)) kg m/sec
And so the momentum acquired by the bigger mass will be p=√(10×5x) Joules>

Answer 2

The law of momentum conservation can be stated as follows. For a collision occurring between object 1 and object 2 in an isolated system, the total momentum of the two objects before the collision is equal to the total momentum of the two objects after the collision.