Question

A stationary body of mass 3kg explodes into three equal parts. Two of the pieces fly off at right angle to each other with the velocities 2im/s and 3jm/s . If the explosion takes place in 10-³sec.find out the average force action on the third piece N.
a) (-2i-3j) 10³
b) (2i+3j) 10³
c) (2i-3j) 10-³
d) none of these​
please answer fast!!! it's urgent!!!!!!!!!!!!

Answer 1

Given :

A stationary body of mass 3kg explodes into three equal parts.

Two of the pieces fly off at right to each other with the velocities of 2i m/s and 3j m/s.

Time interval = 10‾³ s

To Find :

Average force acts on the third piece.

Solution :

❖ Since no net force acts on the system, linear momentum of the system is conserved.

» Initially body was at rest hence initial momentum will be zero.

After explosion body devides into three equal parts.

∴ Mass of each part = 3/3 = 1kg

$\sf:\implies\:\vec{P}_{Initial}=\vec{P}_{Final}$

$\sf:\implies\:0=\vec{P}_1+\vec{P}_2+\vec{P}_3$

$\sf:\implies\:0=m_1\vec{v}_1+m_2\vec{v}_2+m_3\vec{v}_3$

$\sf:\implies\:0=(1)(2\hat{i})+(1)(3\hat{j})+(1)(\vec{v}_3)$

$\bf:\implies\:\vec{v}_3=-2\hat{i}-3\hat{j}$

$\star$ As per newton's second law of motion, force is defined as the rate of change of linear momentum.

$\sf:\implies\:\vec{F}_3=\dfrac{m\vec{v}_3}{t}$

$\sf:\implies\:\vec{F}_3=\dfrac{1(-2\hat{i}-3\hat{j})}{10^{-3}}$

$:\implies\:\underline{\boxed{\bf{\orange{\vec{F}=(-2\hat{i}-3\hat{j})\times10^3\:N}}}}$

Answer 2

Given :

A stationary body of mass 3kg explodes into three equal parts.

Two of the pieces fly off at right to each other with the velocities of 2i m/s and 3j m/s.

Time interval = 10‾³ s

To Find :

Average force acts on the third piece.

Solution :

❖ Since no net force acts on the system, linear momentum of the system is conserved.

» Initially body was at rest hence initial momentum will be zero.

After explosion body devides into three equal parts.

∴ Mass of each part = 3/3 = 1kg

$\sf:\implies\:\vec{P}_{Initial}=\vec{P}_{Final}$

$\sf:\implies\:0=\vec{P}_1+\vec{P}_2+\vec{P}_3$

$\sf:\implies\:0=m_1\vec{v}_1+m_2\vec{v}_2+m_3\vec{v}_3$

$\sf:\implies\:0=(1)(2\hat{i})+(1)(3\hat{j})+(1)(\vec{v}_3)$

$\bf:\implies\:\vec{v}_3=-2\hat{i}-3\hat{j}$

$\star$ As per newton's second law of motion, force is defined as the rate of change of linear momentum.

$\sf:\implies\:\vec{F}_3=\dfrac{m\vec{v}_3}{t}$

$\sf:\implies\:\vec{F}_3=\dfrac{1(-2\hat{i}-3\hat{j})}{10^{-3}}$

$:\implies\:\underline{\boxed{\bf{\orange{\vec{F}=(-2\hat{i}-3\hat{j})\times10^3\:N}}}}$