Kinematics 5 Question 20

19. Particles P and Q of mass 20g and 40g respectively are simultaneously projected from points A and B on the ground. The initial velocities of P and Q make 45 and 135 angles respectively with the horizontal AB as shown in the figure. Each particle has an initial speed of 49m/s. The separation AB is 245m.

Both particles travel in the same vertical plane and undergo a collision. After the collision, P retraces its path. (a) Determine the position Q where it hits the ground. (b) How much time after the collision does the particle Q take to reach the ground? (Take g=9.8m/s2 ).

(1982,8M

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Answer:

Correct Answer: 19. Just midway between A and B,3.53s

Solution:

  1. (a) Range of both the particles is

R=u2sin2θg=(49)2sin909.8

By symmetry we can say that they will collide at highest point.

Just before collision Just after collision

Let v be the velocity of Q just after collision. Then, from conservation of linear momentum, we have

20(ucos45)40(ucos45)=40(v)20(ucos45)v=0

i.e. particle Q comes to rest. So, particle Q will fall vertically downwards and will strike just midway between A and B.

(b) Maximum height,

H=u2sin2θ2g=(49)2sin2452×9.8=61.25m

Therefore, time taken by Q to reach the ground,

t=2Hg=2×61.259.8=3.53s



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