Question

A $4.0\text{kg}$ mass sliding on a frictionless horizontal surface explodes into two equal parts, one moving with $3.0\text{m/s}$ due north and the other with $5.0\text{m/s}$ due ${30}^{\circ }$ north of east. What is the original speed of the mass?

• A. $2.5\text{m/s}$
• B. $3.5\text{m/s}$
• C. $4.5\text{m/s}$
• D. $6.5\text{m/s}$

Answer 1

The correct option is B $3.5\text{m/s}$

Let $u$ be the initial velocity of mass.
After explosion,


On applying the law of conservation of linear momentum along $x-$ axis before and after explosion,
$4\times u_x=2\times 5\cos {30}^{\circ }$
$\Rightarrow u_x=\frac{5\sqrt{3}}{4}\text{m/s}$

On applying the law of conservation of linear momentum along $y-$ axis before and after explosion.
$4\times u_y=2\times 5\sin {30}^{\circ }+2\times 3$
$\Rightarrow u_y=\frac{11}{4}\text{m/s}$

So, velocity of mass before explosion,
$\overrightarrow{u}=(\frac{5\sqrt{3}}{4}\hat{i}+\frac{11}{4}\hat{j})\text{m/s}$

Now, the magnitude of velocity,
$u=\sqrt{{\left(\frac{5\sqrt{3}}{4}\right)}^2+{\left(\frac{11}{4}\right)}^2}$
$\Rightarrow u=3.5\text{m/s}$