Particle of mass 1 kg is thrown vertically upwards with speed 100 m/s-1. After 5 s it explodes into two parts. One part of mass 400 g comes back with speed 25 m/s-1, what is the speed of other part just after explosion?
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Answer:
I think it would be 66.67m/s
Explanation:
Speed of 1kg particle after 5s = 50m/s (using 1 equation of motion)
Now
Initial momentum=m×v= 1×50 = 50 kg m/s
According to law of conservation of momentum
M1V1 = M2V2
Since particular is split into two parts
∴ M2V2 = M2'V2' + M2*V2*
Given
M2'=0.4kg
V2'=25 m/s
From law conservation of mass
M2*=0.6kg
V2*=?
∴ 50 = 0.4×25 + 0.6×V2*
50 = 10+ 0.6V2*
V2* = 40/0.6
V2* = 66.67m/s
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