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# physics momentim questions

Free essay example:

1.  Two billiard balls each of mass 80gm moving in opposite directions with a speed of 36km/hr collide and rebound with the same velocity. Find the change in momentum imparted to each ball due to each other?

Solution :

Change in momentum  =mv- (-mv)

=2mv

=2  80  10ˉ³  10

=1600  10ˉ³

=1.6kgm/sec

1. A ball of mass 200gm falls from a height of 5m and rebounds to a height 1.25m.Find the change in momentum and the average force between the ball and the ground if the time during which they are in contact is 0.2sec? g=10m/sec².

Solution :

Velocity on striking the ground, V=√(2gh)

=√(2  10  5)

=10m/sec

Velocity of rebounding V´=√(2gh´)

=√(2  10  1.25)

=5m/sec

Change in momentum =mv- (-mv´)

=mv+mv´

=m(v+v´)

=((200  10ˉ³)  (10+5))

=(0.2  15)

=3kgm/sec

Force  time = 3kgm/sec

F  0.2 sec= 3kg/sec

So F =  F= 15N

1. A cricket ball of mass 150g moving with a speed 12m/s is hit by bat so that the ball is turned back with a velocity 20m/sec. Calculate  the change in momentum and force applied by the bat if the ball is in contact with the bat for a time of 1  10ˉ²sec ?

Solution :

Change in momentum =mv- mu

=m(v-u)

=((150  10ˉ³)  (-20-12))

=-4.8kgm/sec

Force  time =-4.8kgm/sec

F =  = -  = -4.8  10² N

1. A ball of mass 0.2kg moves with a velocity 20m/sec is brought to rest by a player in 0.1sec.find the change in momentum and the average force applied to bring the ball to rest ?

Solution :

Change in momentum =mv- mu

=m(v-u)

=(0.2  (0-20))

=4.0kgm/sec

F = 4.0/0.1sec

= 40 N

1. A body falls from a height of 100cm and rebounds to a height 50cm.Then change in momentum is 0.3788Ns.If the force exerted by the body on the ground is 3.78N find the mass of the body and the time of contact of it with the ground ?

Solution :

Velocity on striking the ground , V1=√(2gh´)

=√(2  9.8  1)

=4.427m/sec

Velocity of rebounding , V2 = √(2gh´´)

=√(2  9.8  (50  10ˉ²))

=3.13m/sec

Change in momentum=  mv- mu

=m(v-u)

=m(v- (-u))

=m(v+u)

0.3778  =m(4.427+3.13)

m=  0.050kg

Force  time =0.3778  kgm/sec

T = 0.3788/3.78

T=0.100 sec

1. A  batsman hits back a ball straight in  direction of the bowler without changing the initial speed of 12m/sec. If the change in momentum imparted to the ball is 3.6Ns from batsman to bowler find the mass of the ball  if the ball is in contact with the ball for 0.2sec .  find the force exerted by the batsman?

Solution :

Change in momentum =m(v- u)

3.6 = -24m

m= 3.6/24

m= 0.15kg

Force  time = Change in momentum

F  0.2sec = 3.6

F = 3.6/0.2

=18 N

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