Kinematics Question

Question

A ball of mass 100 g is projected vertically upwards from the ground with velocity 49 m/s. At the same time other identical ball is dropped from a height of 98m to fall freely along the same path as followed by the first ball. After some time two balls collide and stick together and finally fall together. Find the time of flight of the masses.

Solution

We will first find where and when the two balls collide. Let us assume that the balls collide at time t after they have been set into motion. At this instant t when two balls collide they are at the same height h from the ground as shown below in the figure.

The height of the first ball after t seconds = 49t-0.5(9.8t²) = 4.9t(20-t)

Height of second ball after t secinds = 98 - downwards distance moved by it in t seconds

                                                      =98-0.5t²=4.9(20-t²)

therefore, 4.9t(20-t)=4.9(20-t²)

or, 10t-t²=20-t² or t=2s

The ball thus collides 2s after the start of their motion. Their velocities at this instance are

ball 1 : v₁= (49-98×2)m/s = 29.4 m/s directed upwards

ball 2 : v₂=(0+9.8×2)m/s = 19.6 m/s directed downwards

If v is the velocity of the combined mass of two balls after they stick togather due to their collision then from law of conservation of momentum

200×v=100×29.4-100×19.6

v=4.9m/s

The joint mass thus moves upwards , after collision with a velocity of 4.9 m/s. Its height above the ground at this instant is (consider the position of either of the balls)

(98-0.5×9.8×2²)m=78.4m

We can now find the time t' taken by this joined mass of the balls to reach the ground. For this joined mass we have

u=4.9m/s , s=78.4m , a=-g = -9.8m/s²

-78.4=4.9t'+0.5(-9.8)t²

t'²-t'-16=0

Solving the equation for t' using formula for quadratic equations and leaving out the negative solution we get t'=4.532 s

The joint mass thus takes 4.53 s to fall to the ground. Since the balls collide 2s after they started their motion ,

the total time of flight is (2+4.53) s = 6.53 s