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A bomb at rest explodes

When a bomb at rest explodes into some fragments, we can find the velocity of it’s fragments using Law of Conservation of Linear Momentum.

Suppose a bomb at rest explodes into two pieces with their masses in the ratio 3 : 1. Let :

3 m = mass of first pieces, m = mass of second pieces

v₁ = velocity of first pieces, v₂ = velocity of second pieces

Now before explosion, the bomb was at rest, so initial linear momentum, p_i = 0.

Total linear momentum after explosion, p_f = 3 m × v₁ + m × v₂

As the explosion is due to internal forces, applying Law of Conservation of Linear Momentum :

p_i = p_f

0 = 3 m × v₁ + m × v₂

⇒ v₂= – 3 v₁

Hence smaller mass moves in opposite direction with velocity 3v₁.

Example 1

A bomb at rest splits into three identical parts. If two pieces are moving at right angles to each other with same velocity v m/s, the velocity of third fragment is

(A) v/√2

(B) v/2

(D) v√2

Solution :-

Before splitting, initial linear momentum, p_i = 0.

Let the three identical masses are m/3 each.

After splitting, final linear momentum, $\displaystyle {{\overrightarrow{p}}_{f}}=\frac{m}{3}(v\widehat{i})+\frac{m}{3}(v\widehat{j})+\frac{m}{3}(\overrightarrow{v'})$

From the Law of Conservation of Linear Momentum,

p_f = p_i

$\displaystyle \frac{m}{3}(v\widehat{i})+\frac{m}{3}(v\widehat{j})+\frac{m}{3}(\overrightarrow{v'})=0$

$\displaystyle v\widehat{i}+v\widehat{j}+\overrightarrow{v'}=0$

$\displaystyle \overrightarrow{v'}=-(v\widehat{i}+v\widehat{j})$

$\displaystyle v'=\sqrt{2}v$

Example 2

A bomb at rest explodes into three fragments. Two fragments fly off at right angles to each other. If the velocity of 1 kg fragment is 12 m/s and 2 kg is 8 m/s, which are at right angles and velocity of third fragment is 40 m/s, calculate the mass of third fragment.

Solution :-

Applying Conservation of Linear Momentum,

p_f = p_i

$\displaystyle {{\overrightarrow{p}}_{1}}+{{\overrightarrow{p}}_{2}}+{{\overrightarrow{p}}_{3}}=0$

$\displaystyle \Rightarrow {{\overrightarrow{p}}_{3}}=-({{\overrightarrow{p}}_{1}}+{{\overrightarrow{p}}_{2}})$

$\displaystyle \Rightarrow {{p}_{3}}=\sqrt{p_{1}^{2}+p_{2}^{2}+2{{p}_{1}}{{p}_{2}}\cos {{90}^{\circ }}}$

$\displaystyle \Rightarrow {{p}_{3}}=\sqrt{p_{1}^{2}+p_{2}^{2}}$

$\displaystyle {{m}_{3}}{{v}_{3}}=\sqrt{m_{_{1}}^{2}v_{1}^{2}+m_{_{2}}^{2}v_{2}^{2}}$

$\displaystyle {{m}_{3}}=\frac{\sqrt{m_{_{1}}^{2}v_{1}^{2}+m_{_{2}}^{2}v_{2}^{2}}}{{{v}_{3}}}$

$\displaystyle {{m}_{3}}=\frac{\sqrt{{{(1\times 12)}^{2}}+{{(2\times 8)}^{2}}}}{40}$

$\displaystyle {{m}_{3}}=\frac{\sqrt{144+256}}{40}=0.5kg$

Example 3

A bomb at rest explodes into three equal fragments. Two fragments fly off at right angles with velocity $12 m/s$ and $9 m/s$ . If the explosion process occurs in $0.1 s$ , find out the average force acting on the third particle, if total mass of the bomb is $12 kg$ .

Solution :-

Mass of each fragment = $\frac{12/}{3} = 4 kg$ .

As two fragments fly off at right angles, let one moves along x-axis and another along y-axis. Let the velocity of third fragment be $\displaystyle \overrightarrow{v}$ .

Applying Conservation of Linear Momentum,

p_f = p_i

$\displaystyle \left( 4\times 12\widehat{i} \right)+\left( 4\times 9\widehat{j} \right)+4\overrightarrow{v}=0$

∴ Speed of third particle, $\displaystyle \left| \overrightarrow{v} \right|=\sqrt{{{12}^{2}}+{{9}^{2}}}=15m/s$

Average force acting on the third particle = change in momentum/time

$\displaystyle \Rightarrow F=\frac{4\times 15-0}{0.1}=600N$

A bomb at rest explodes | Explosion of a bomb at rest

A bomb at rest explodes

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