NEET Class neet Answered
A project file of mass 8 kg is fired at 25 m per second at 37 degree from horizontal it explodes at highest point into 3 parts of ratio 1:3:5 smallest part retraces it's path while second smaller drops vertically find velocity of biggest mass and also find explosion energy
Asked by harshs2580 | 04 Nov, 2022, 08:45: AM
Expert Answer
When projectile reaches the maximum height , its velocity in vertical direction is zero .
Velocity of projectile in horizontal direction is ( 25 × cos37 ) m/s
Hence projectile has momentum just before explosion only in horizontal direction.
Horizontal direction momentum just before explosion = mass × velocity
Horizontal direction momentum just before explosion = 8 × ( 25 × cos37 ) m/s = 160 kg m/s
After explosion projectile mass is fragmented in the ratio 1 : 3 : 5 .
Mass of smallest fragment = 8 × (1/9) kg = ( 8/9 ) kg
If smallest fragment retraces its path after explosion , then its horizontal velocity just after explosion is ( 25 × cos37 ) m/s.
Vertical direction velocity of smallest fragment is zero . Hence smallest fragment has momentum only in horizontal direction
Horizontal direction momentum of smallest fragment = ( 8/9 ) × ( 25 × cos37 ) = 17.78 kg m/s
If next smallest segment is just dropped after explosion, then its velocity is zero after explosion.
Mass of largest fragment = ( 5 / 9 ) × 8 kg = ( 40/9 ) kg
Initial momentum before explosion is only in horizontal direction .
Momentum of smallest fragment is only in horizontal direction and momentum of next smallest fragment is zero.
Hence , largest fragment has momentum only in horizontal direction
Next smallest segment does not have any momentum.
Momentum of largest fragment after explosion is calculated from conservation of total momentum
Momentum of largest fragment after explosion = ( 160 - 17.78 ) kg m s-1 = 142.22 m/s
Horizontal direction velocity of largest fragment = momentum / mass = 142.22 / ( 40/9) = 32 m/s
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Initial kinetic energy just before explosion = (1/2) × m v2 = 0.5 × 8 × ( 25 × cos37)2 J
Initial kinetic energy just before explosion = 1600 J
Kinetic energy just after explosion = (1/2) ( 8/9) ( 25 × cos37)2 + (1/2) (40/9) (32)2
Kinetic energy just after explosion =2453 J
Energy released in explosion = ( 2453 -1600 ) J = 853 J
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