Question

A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g. what is the velocity acquired by the block?​

Answer 1

Answer:

$\rm 4.39 {ms}^{ - 1}$

Explanation:

Given,

A bullet of mass 10g with a 400m/s gets embedded in a freely suspended wooden block of mass 900g.

What is the velocity acquired by the block?

Let's say that the velocity acquired by the block is $v$

We know that,

$\bullet \:\: p = mv$

Where,

• p is used to denote the momentum.
• m is the mass and v is the velocity of the object.

Total momenta of the bullet before collision,

$= {\rm m_1}u$

• $u$ denotes the initial velocity of the bullet.

$= (10g)\ast(400ms^{-1})$

$= (0.01kg) \ast (400ms^{-1}) \: \: \: \: \rm[ \because On \: conversion]$

$\rm = 4 kg ms^{-1}$

Total momenta of the bullet and the wooden block after collision,

$= {\rm (m_1 + m_2)} \ast v$

$= {\rm{ (m_1 + m_2)}} \ast v$

$= {\rm{ (0.01kg + 900g)}} \ast v$

$= {\rm{ (0.01kg + 0.9kg)}} \ast v \: \: \: \: \rm[ \because On \: conversion]$

$= {\rm{ (0.91kg)}} \ast v$

$= 0.91v \: \rm kg \: m {s}^{ - 1}$

According to the third law of motion,

Total momenta before collision is equal to Total momenta after collision.

$\implies \: 0.91v = 4$

$\implies \: v = \dfrac{4}{0.91}$

$\therefore \boldsymbol{v} = 4.39$

$\underline{ \sf \therefore \: required \: answer = \rm 4.39 {ms}^{ - 1} }$

Answer 2

CORRECT QUESTION :-

• A bullet of mass 10 g moving with a velocity of 400 m/s gets embedded in a freely suspended wooden block of mass 900g. what is the velocity acquired by the block?

GIVEN :-

• mass of bullet = 10 g = 0.01 kg

• velocity of bullet = 400 ms -¹

• mass of block = 0.9 kg

TO FIND :-

• velocity of the block = ?

SOLUTION :-

momentum of the bullet =

mass of bullet × velocity of bullet

= 0.01 x 400

= 4kg ms -¹

This 10 g bullet embedded in 900 g block so

the total mass of the bock

= 10 + 900

= 0.91 kg

momentum of (wooden block + bullet)= mass of (wooden block + bullet) × velocity (wooden block + bullet)

= 0.91 × v

Law of conservation of linear momentum

momentum of the bullet momentum of

(wooden block + bullet)

4 = 0.91 x v

v = 4 / 0.91

v = 4.4 ms -¹

Hence the answer is 4.4m/s