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Question

A projectile is fired at speed of 502 m/s at an angle of 45 from the horizontal. At t=5 s, the projectile breaks into two parts of mass ratio 1:4 and the lighter piece comes to rest immediately. Find the distance from the launching point to the point where the heavier piece lands.

A
250 m
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B
750 m
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C
500 m
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D
562.5 m
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Solution

The correct option is D 562.5 m

Horizontal component of velocity, ux=ucosθ=ux=502cos45=50 m/s

Lighter piece comes to rest after 5 s.
Therefore, Horizontal distance travelled by lighter piece
=vxt=50×5=250 m

Taking the point of projection as the reference point.
​​xCOM= Range of projectile =R
R=2u2sinθcosθg=2×(502)2×1/2×1/210=500 m

xCOM=m1x1+m2x2m1+m2=m×250+4m×x24m+m=250+4×x25
500=250+4×x254×x2+250=5×500=2500
x2=25002504=22504=562.5 m

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