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Question

A projectile of mass 4M is fired at a speed of 100 m/s at an angle of 37 from the horizontal. At the highest point, the projectile breaks into two parts with mass M and 3M. Mass M falls down immediately after the explosion and mass 3M lands at some horizontal distance from mass M. Choose the correct statement. (Take g=10 m/s2)

A
Time taken by 4M to reach the highest point and time taken by 3M to land on the ground from the highest point will be the same.
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B
Time taken by 4M to reach the highest point and time taken by 3M to land on the ground from the highest point will be different.
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C
Distance covered by 4M along the horizontal ground and distance covered by 3M along the horizontal ground will be the same.
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D
None of the above
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Solution

The correct option is A Time taken by 4M to reach the highest point and time taken by 3M to land on the ground from the highest point will be the same.
For 4M, the components of velocity are:
ux=100cos37=80 m/s
uy=100sin37=60 m/s
Time taken by 4M to reach maximum height
= Time of ascent ta=uyg=6010=6 s

Horizontal component of initial momentum Pix will be same for both fragments 4M and M at maximum height, since force in horizontal direction Fx=0


M is at rest after explosion its horizontal velocity vx=0
Let 3M moves with vx in horizontal after explosion, and vy=0

From momentum conservation:
4M×ux=(M×0)+(3M×vx)
or, 4M×80=0+3Mvx
vx=3203 m/s

vx>ux, 3M fragment will cover a greater horizontal distance as compared to unbroken 4M mass during descent time.
Hence, option (c) is incorrect.

Time taken by 3M to reach the ground from highest point will be time of descent (td)
td for 3M=ta for 4M

vy (final velocity in y-direction) will be the same for both the fragments, which determines time of ascend/descend.
td=ta=6 s
Hence option (a) is correct and option (b) is incorrect.

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