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

Verified by Experts

Mass of bullet , m = 0.012 kg
Speed of bulet , `v = 70 " m "s^(-1)`
Mass of wooden block of , M = 0.4 kg Let V be the velocity of system consisting of block and the bullet having a mass `M+m`
According to the law of conservation of momentm , `mv=(m+M)V`
`0.012xx70 =(0.012+0.4)V`
`V = (0.012 xx70)/(0.412) =2.04 " m"s^(-1)`
Applyng `v^(2)-u^(2)=2gh `to the system of block and bullet
`(2.04)^(2)=2xx9.8 xxh`
Doubtnut Promotions Banner Mobile Dark
|

Topper's Solved these Questions

  • WORK, ENERGY AND POWER

    MODERN PUBLICATION|Exercise NCERT FILE (Exemplar Problems Objective Questions)|21 Videos

  • WORK, ENERGY AND POWER

    MODERN PUBLICATION|Exercise NCERT FILE (NCERT Exemplar Problems (Very Short Answer Type Questions))|12 Videos

  • WORK, ENERGY AND POWER

    MODERN PUBLICATION|Exercise NCERT FILE (NCERT Textbook Exercise)|23 Videos

  • WAVES

    MODERN PUBLICATION|Exercise CHAPTER PRACTICE TEST|14 Videos

Similar Questions

Explore conceptually related problems

A bullet of mass 10 kg moving horizontally at a speed 140 m//s strikes a block of mass 100 g attached to a string like a simple pendulum. The bullet penetrates the block and emerges on the other side. If the block rises by 80 cm , find final velocity of bullet.

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`

  • A bullet of mass 50 grams going at a speed of 200 ms^(-1) strikes a wood block of mass 950 gm and gets embedded in it. The velocity of the block after the impact is

    A
    `5 ms^(-1)`
    B
    `10 ms^(-1)`
    C
    `20 ms^(-1)`
    D
    `50 ms^(-1)`

  • 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
    `15ms^(-1)`
    B
    `20ms^(-1)`
    C
    `100ms^(-1)`
    D
    `50ms^(-1)`

    • Similar Questions

    Explore conceptually related problems

    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 0.06 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 10 g moving horizontal of 400 ms^(-1) strikes a wooden block of suspended by a light inextensible string a result, the center of gravity of the block: vertical distance of 10 cm. The speed of the emerges out horizontally from the block will

    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. If the lighter block comes to rest after collision, then the speed of the heavier body 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

    Home
    Profile