A projectile of mass 50 kg is shot vertically upwards with a velocity of 100 m/sec. After 5 sec the projectile explodes into two fragments, one of which(mass 20 kg) travels up with 150 m/s. what is the velocity of other fragment at that moment?
Ans/ 15 m/s.

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Solution

Velocity of the 50 kg mass after 5 second = 100 - 5g = 50 m/s
Let the velocity of the other fragment be v.
Applying law of conservation of momentum we get that
50 x 50 = 20 x 150 - (50 - 20)v
Or, 2500 = 3000 - 30v
Therefore, v = 500/30 = 50/3 ms^-1 in the vertically downward direction

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A projectile of mass 50 kg is shot vertically upwards with a velocity of 100 m/sec. After 5 sec the projectile explodes into two fragments, one of which(mass 20 kg) travels up with 150 m/s. what is the velocity of other fragment at that moment?Ans/ 15 m/s.

Velocity of the 50 kg mass after 5 second = 100 - 5g = 50 m/sLet the velocity of the other fragment be v.Applying law of conservation of momentum we get that50 x 50 = 20 x 150 - (50 - 20)vOr, 2500 = 3000 - 30vTherefore, v = 500/30 = 50/3 ms-1 in the vertically downward direction

A projectile of mass 50 kg is shot vertically upwards with a velocity of 100 m/sec. After 5 sec the projectile explodes into two fragments, one of which(mass 20 kg) travels up with 150 m/s. what is the velocity of other fragment at that moment?
Ans/ 15 m/s.

Velocity of the 50 kg mass after 5 second = 100 - 5g = 50 m/s
Let the velocity of the other fragment be v.
Applying law of conservation of momentum we get that
50 x 50 = 20 x 150 - (50 - 20)v
Or, 2500 = 3000 - 30v
Therefore, v = 500/30 = 50/3 ms^-1 in the vertically downward direction