1.

When a projectile is moving at 60 m s^(-1) at the highest point of its trajectory, it explodes into two equal parts. One part moves vertically up with a velocity of 50 ms^(-1) . The magnitude of velocity of other part is:

Answer»

`50ms^(-1)`
`60ms^(-1)`
`120ms^(-1)`
`130ms^(-1)`

Solution :Apply law of conservation of momentum Let `v_(x)` be the horizontal velocity of the other PART.
Applying momentum conservation along X-axis, we get
`mv_(x)=(2m)60`
or `v_(x)=120ms^(-1)`
Applying momentum conservation along Y-axis, we get
`mv_(y)=m(50)`
or `v_(y)=50ms^(-1)`
Velocity of the other part, v
`sqrt(120^(2)+50^(2))MS^(-1)`


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