Kinematics Question


A ball of mass 100 g is projected vertically upward from the ground with avelocity of 49m/s. At the same time another identical ball is dropped from aheight of 98 m to fall freely along the same path as followed by the firstball. After some time ,the two balls collide and stick together and finallyfall together .


i is the unit vector along upward direction

1) Find the time at which balls collide in air

a) 2.1 sec

b) 1.2 sec

c)1 sec

d) 2 sec

2) Find the velocity vector of the combined mass just after collision

a) 4.8i

b) 3i


d) 4.9i

3) Find the time of the flight of masses

a) 3sec

b) 4 sec

c) 6.53 sec

d) 7 sec


a) We will first find where and when the two balls collide. Let us assumethat the balls collide at time t after they have been set into motion. At thisinstant t when two balls collide they are at the same height h from the groundas 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 itin t seconds


therefore, 4.9t(20-t)=4.9(20-t2)

or, 10t-t2=20-t2 or t=2s

b) The ball thus collides 2s after the start of their motion. Theirvelocities at this instance are

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

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

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


v=4.9m/s upward direction =4.9 i

c) The joint mass thus moves upwards , after collision with a velocity of4.9 m/s. Its height above the ground at this instant is (consider the positionof either of the balls)


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

u=4.9m/s , s=78.4m , a=-g = -9.8m/s2



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

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