Question

2 bodies a and b are thrown up with 100m/sec from the same point with an interval of 2sec find when and where they will collide and at what velocities they will colide
pls help

Answer 1

Explanation:

PHYSICS

Two balls of equal masses are thrown upwards, along the same vertical direction at an interval of 2 seconds, with the same initial velocity of 40m/s. Then these collide at a height of (Take g=10m/s

2

).

December 27, 2019avatar

Ranjan Abhista

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ANSWER

Let the two balls collide t sec after the first ball is thrown, and let h be the height at which they collide.

For the first ball:

h=40t−

2

1

gt

2

And for the second ball:

h=40(t−2)−

2

1

g(t−2)

2

40t−

2

1

gt

2

=40(t−2)−

2

1

g(t−2)

2

2gt=80+20=100

t=5 sec

∴h=40×5−

2

1

×10×5

2

=75m

Hence, option B is correct.

December 27, 2019

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Toppr

Answer 2

Displacement of 1st mass = s1 = 39.2 t - 1/2 * 9.8 * t²  =  4.9 (8 t - t²)

Displacement of 2nd mass = s2 = 39.2 (t-2) - 1/2 * 9.8 * (t-2)²

      �              = 4.9 [ 8(t-2) - (t-2)² ] = 4.9 [ 8 t - 16 - t² + 4 t - 4 ]

                   = 4.9 [ 8 t - t² +4 t - 20 ]

s1 = s2    =>       4 t - 20 =0                 =>  t = 5 sec

s1 = height at which they collide = 4.9 ( 8*5 - 5²) = 4.9 * 15 = 73.5 meters