Question

A particle of mass $m=4\text{kg}$ is projected with a speed of $u=10\text{m/s}$ at an angle of ${45}^{\circ }$ with the horizontal. The particle explodes in mid air when it is at the maximum height. One component weighing $1\text{kg}$ comes to rest after the explosion and the other component of mass $3\text{kg}$ moves away. Find the distance where the $3\text{kg}$ mass component strikes the ground from the initial position (projection point). [Take $g=10{\text{m/s}}^2$]

• A. $15\text{m}$
• B. $\frac{35}{3}\text{m}$
• C. $\frac{20}{3}\text{m}$
• D. $25\text{m}$

Answer 1

The correct option is B $\frac{35}{3}\text{m}$


Time taken by the particle to reach the highest point
$t=\frac{u\sin {45}^{\circ }}{g}=\frac{10\text{sin}{45}^{\circ }}{10}=\frac{1}{\sqrt{2}}\text{sec}$
Distance covered in horizontal direction:
$S_x=u_{x}t=10\cos {45}^{\circ }\times t$
$=\frac{10\cos {45}^{\circ }}{\sqrt{2}}=5\text{m}$

No external force is acting on the system in $x$ direction. So, momentum will be conserved in $x$ direction.
Let the velocity of the $3\text{kg}$ component be $v^{\prime }$.
$p_i=p_f$ along $x$ direction
$\Rightarrow 4\left(u\cos {45}^{\circ }\right)=1\times \left(0\right)+3v^{{}^{\prime }}$
$\Rightarrow v^{{}^{\prime }}=\frac{4}{3}\times 10\times \frac{1}{\sqrt{2}}=\frac{20\sqrt{2}}{3}\text{m/s}$

Time taken by the $3\text{kg}$ particle to reach the ground will be same of the time taken by the particle to reach at top. (since vertical velocity is not affected)
Let $S_x^{{}^{\prime }}$ = distance travelled (by $3\text{kg}$ particle)
$S_x^{{}^{\prime }}=v^{{}^{\prime }}t+\frac{1}{2}{a}_x{t}^2(a_x=0)$
$\Rightarrow S_x^{{}^{\prime }}=\frac{20\sqrt{2}}{3}\times \frac{1}{\sqrt{2}}=\frac{20}{3}\text{m}$
Thus, distance from the projection point = $5+\frac{20}{3}=\frac{35}{3}\text{m}$