6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
6.24 A bullet of mass 0.012 kg and horizontal speed 70 m s–1 strikes a block of wood of mass 0.4 kg and instantly comes to rest with respect to the block. The block is suspended from the ceiling by means of thin wires. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.
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1 Answer
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6.24 Mass of the bullet, = 0.012 kg
Initial speed of the bullet, u = 70 m/s
Mass of the wooden block , = 0.4 kg
Initial speed of the wooden block = 0
Let's assume, final speed of the bullet = v
Applying the law of conservation of momentum
Hence v = ( = 2.04 m/s
Let h be the height by which the block rise. Applying law of conservation of energy
Potential energy of the combined bullet + block = Kinetic energy of the combination
(1/2)
h = /2g = 0.212 m
The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination
= (1/2) - (1/2)
= (1/2
...more
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(b) conserving energy between “O” ans ”A”

Ui + Ki = Uf + Kf
0+1/2mv2= mgh + 1/2mv’
(v’)2=v2-2gh = v’= ……….1
Let speed after emerging be v1 then
=1/2mv12=1/2[1/2mv’2]
1/2m(v1)2=1/4m(v’)2=1/4m[v2-2gh]
V1= ………….2
From eqn 1 and 2
So v1 = v’/ =v2(v’/2)
v1>v’/2
hence after emerging from the target velocity of the bullet is more than half of its earlier velocity v’
(d) as the velocity of the bullet changes to v’ which is less than v’ hence , path
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(b, d) When a man of mass m climbs up the staircases of height L, work done by the gravitational force on the man is mgl work done by internal muscular forces will be mgL as the change in kinetic is almost zero.
Hence total work done =-mgL + mgL=0
As the point of application of the contact forces does not move hence work done by reaction forces will be zero.
This is a multiple choice answer as classified in NCERT Exemplar
(c) m =150g =3/20kg
Time of contact =0.001s
U=126km/h=
V= -35m/s
Change in momentum of the ball = m (v-u)=
=21/2
F= dp/dt=- = - 1.05
Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.
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