Home
Class 11
PHYSICS
A bullet of mass 0.012 kg and horizontal...

A bullet of mass 0.012 kg and horizontal speed `70ms^(-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 thin wire. Calculate the height to which the block rises. Also, estimate the amount of heat produced in the block.

Text Solution

AI Generated Solution

To solve the problem step by step, we will follow these steps: ### Step 1: Calculate the initial momentum of the bullet and the block. The initial momentum of the bullet can be calculated using the formula: \[ \text{Initial Momentum} = \text{mass of bullet} \times \text{velocity of bullet} \] Given: ...
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • WORK, ENERGY AND POWER

    NCERT ENGLISH|Exercise EXERCISE|30 Videos

  • WAVES

    NCERT ENGLISH|Exercise EXERCISE|27 Videos

Similar Questions

Explore conceptually related problems

A bullet of mass 0.01 kg and travelling at a speed of 500 ms^(-1) strikes a block of mass 2 kg which is suspended by a string of length 5 m . The centre of gravity of the block is found to raise a vertical distance of 0.2 m . What is the speed of the bullet after it emerges from the block?

A bullet of mass 0.02 kg travelling horizontally with velocity 250 ms^(-1) strikes a block of wood of mass 0.23 kg which rests on a rough horizontal surface. After the impact, the block and bullet move together and come to rest after travelling a distance of 40m. The coefficient of sliding friction of the rough surface is ( g = 9.8 ms^(_2))

Knowledge Check

  • A bullet of mass m moving horizontally with a velocity v strikes a block of wood of mass M and gets embedded in the block. The block is suspended from the ceiling by a massless string. The height to which block rises is

    A
    `v^2/(2g)(m/(M+m))^2`
    B
    `v^2/(2g)((M+m)/(m))^2`
    C
    `v^2/(2g)(m/M)^2`
    D
    `v^2/(2g)(M/m)^2`

    • Similar Questions

    Explore conceptually related problems

    A block of mass 1 kg is placed on a horizontal surface . The reaction force to the weight of the block is

    A block of mass 1 kg moving with a speed of 4 ms^(-1) , collides with another block of mass 2 kg which is at rest. The lighter block comes to rest after collision. The loss in KE of the system is

    A bullet of mass 0.04 kg moving with a speed of 90 ms^(-1) enters a heavy wooden block and is stooped after a distance of 60 cm . What is the average resistive force exerted by the block on the bullet ?

    A bullet of mass 0.005 kg moving with a speed of 200 ms^(-1) enters a heavy wooden block and is stopped after a distance of 50 cm. What is the average force exerted by the block on the bullet?

    A bullet of mass 0.06 kg moving with a speed of 90ms^(-1) enters a heavy wooden block and it stopped after a distance of 60 cm. What is the average resistive force exerted by the block on the bullet ?

    A bullet of mass 10g travelling horizontally with a velocity of 150 ms^(-1) strikes a stationary wooden block and come to rest in 0.03 s . Calculate the distance of penetration of the bullet into the block. Also, Calculate the magnitude of the force exerted by the wooden block on the bullet,

    A bullet of mass 0.01 kg and travelling at a speed of 500 m/s strikes a block of mass 2 kg which is suspended by a string of length 5m. The center of gravity of the block is found to rise a vertical distance of 0.1m, figure. What is the speed of the bullet after it emerges from the block ? (g=9.8 m//s^2) .

    Home
    Profile