Kinematics Question

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 .

g=9.8m/s2

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

c)-4.9i

d) 4.9i

3) Find the time of the flight of masses

a) 3sec

b) 4 sec

c) 6.53 sec

d) 7 sec

Solution:

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

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

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

200×v=100×29.4-100×19.6

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)

(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 reachthe 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

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