Question

A projectile of mass $4M$ is fired at a speed of $100\text{m/s}$ at an angle of ${37}^{\circ }$ from the horizontal. At the highest point, the projectile breaks into two parts with mass $M$ and $3M$. Mass $M$ falls down immediately after the explosion and mass $3M$ lands at some horizontal distance from mass $M$. Choose the correct statement. (Take $g=10{\text{m/s}}^2$)

• A. Time taken by $4M$ to reach the highest point and time taken by $3M$ to land on the ground from the highest point will be the same.
• B. Time taken by $4M$ to reach the highest point and time taken by $3M$ to land on the ground from the highest point will be different.
• C. Distance covered by $4M$ along the horizontal ground and distance covered by $3M$ along the horizontal ground will be the same.
• D. None of the above

Answer 1

The correct option is A Time taken by $4M$ to reach the highest point and time taken by $3M$ to land on the ground from the highest point will be the same.

For $4M$, the components of velocity are:
$u_x=100\cos {37}^{\circ }=80\text{m/s}$
$u_y=100\sin {37}^{\circ }=60\text{m/s}$
Time taken by $4M$ to reach maximum height
$=$ Time of ascent $t_a=\frac{u_y}{g}=\frac{60}{10}=6\text{s}$

Horizontal component of initial momentum $P_i{}_x$ will be same for both fragments $4M$ and $M$ at maximum height, since force in horizontal direction $F_x=0$


$\because M$ is at rest after explosion its horizontal velocity $v_x=0$
Let $3M$ moves with $v_x^{\prime }$ in horizontal after explosion, and $v_y^{\prime }=0$

From momentum conservation:
$4M\times u_x=(M\times 0)+(3M\times v_x^{\prime })$
or, $4M\times 80=0+3M{v}_x^{\prime }$
$\therefore v_x^{\prime }=\frac{320}{3}\text{m/s}$

$\because v_x^{\prime }>u_x$, $3M$ fragment will cover a greater horizontal distance as compared to unbroken $4M$ mass during descent time.
Hence, option ($c$) is incorrect.

Time taken by $3M$ to reach the ground from highest point will be time of descent ($t_d$)
$t_d$ for $3M=t_a$ for $4M$

$\because v_y$ (final velocity in y-direction) will be the same for both the fragments, which determines time of ascend/descend.
$\therefore t_d=t_a=6\text{s}$
Hence option $\left(a\right)$ is correct and option $\left(b\right)$ is incorrect.