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Question

A particle of mass m=4 kg is projected with a speed of u=10 m/s at an angle of 45 with the horizontal. The particle explodes in mid air when it is at the maximum height. One component weighing 1 kg comes to rest after the explosion and the other component of mass 3 kg moves away. Find the distance where the 3 kg mass component strikes the ground from the initial position (projection point). [Take g=10 m/s2]

A
15 m
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B
353 m
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C
203 m
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D
25 m
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Solution

The correct option is B 353 m

Time taken by the particle to reach the highest point
t=usin45g=10sin4510=12 sec
Distance covered in horizontal direction:
Sx=uxt=10cos45×t
=10cos452=5 m

No external force is acting on the system in x direction. So, momentum will be conserved in x direction.
Let the velocity of the 3 kg component be v.
pi=pf along x direction
4(ucos45)=1×(0)+3v
v=43×10×12=2023 m/s

Time taken by the 3 kg particle to reach the ground will be same of the time taken by the particle to reach at top. (since vertical velocity is not affected)
Let Sx = distance travelled (by 3 kg particle)
Sx=vt+12axt2 (ax=0)
Sx=2023×12=203 m
Thus, distance from the projection point = 5+203=353 m

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