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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    Answered by

    Payal Gupta | Contributor-Level 10

    4 months ago

    6.24 Mass of the bullet, mb = 0.012 kg

    Initial speed of the bullet, u = 70 m/s

    Mass of the wooden block , mw = 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

    mb*u+mw*0=mb+mwv

    Hence v = ( mb*u)/mb+mw = 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

    mb+mw*g*h= (1/2) *mb+mw*v2

    h = v2 /2g = 0.212 m

    The heat produced = Initial kinetic energy of the bullet – final kinetic energy of the combination

    = (1/2) mb u2 - (1/2) *mb+mw*v2

    = (1/2

    ...more

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(c) When M1 comes in contact with the spring. M1 is retarded by the spring force and M2 is accelerated by the spring force. 
 
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(b) conserving energy between “O” ans ”A”

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This is a multiple choice answer as classified in NCERT Exemplar

(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.

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(c) m =150g =3/20kg

Time of contact =0.001s

U=126km/h= 126 × 1000 60 × 60 = 35 m s

V= -35m/s

Change in momentum of the ball = m (v-u)= 3 20 - 35 - 35 k g m / s

=21/2

F= dp/dt=- 21 / 2 0.001 N = - 1.05 × 10 4 N

Here – negative sign indicates that force will be opposite to the direction of movement of the ball before hitting.

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