Question

A cracker is thrown into the air with a velocity of $10{\text{m}}{{\text{s}}^{ - 1}}$ at an angle of $45^\circ $ with the vertical. When it is at a height of $0 \cdot 5{\text{m}}$ from the ground it explodes into a number of pieces which follow different parabolic paths. Find the velocity of the centre of mass when it is at a height of ${\text{1m}}$ from the ground. (Take $g = 10{\text{m}}{{\text{s}}^{ - 2}}$).
A. $4\sqrt 5 {\text{m}}{{\text{s}}^{ - 1}}$
B. $2\sqrt 5 {\text{m}}{{\text{s}}^{ - 1}}$
C. $4\sqrt 4 {\text{m}}{{\text{s}}^{ - 1}}$
D. ${\text{10m}}{{\text{s}}^{ - 1}}$

Answer 1

Hint: The cracker is thrown upwards with an initial velocity directed at an angle and so the initial velocity will have non-zero components in the x and y directions. As the cracker explodes on its own, the centre of mass of the system will remain unchanged and move with the final velocity of the cracker which can be determined by Newton’s third equation of motion.

Formulas used:
->Newton’s third equation of motion gives the final velocity of an object as ${v^2} = {u^2} + 2as$ where $u$ is the initial velocity of the object, $a$ is its acceleration and $s$ is the displacement of the object.
->The net velocity of a body is given by, ${v_{net}} = \sqrt {{v_x}^2 + {v_y}^2} $ where ${v_x}$and ${v_y}$ are the x-component and y-component of the velocity of the object.

Complete step-by-step solution:
->Step 1: List the parameters mentioned in the question.
The initial velocity of the cracker is given to be $u = 10{\text{m}}{{\text{s}}^{ - 1}}$ and is directed at an angle of $\theta = 45^\circ $ with the vertical.
So the x-component and y-component of the initial velocity will be ${u_x} = u\cos \theta = \dfrac{{10}}{{\sqrt 2 }}{\text{m}}{{\text{s}}^{ - 1}}$ and ${u_y} = u\sin \theta = \dfrac{{10}}{{\sqrt 2 }}{\text{m}}{{\text{s}}^{ - 1}}$
We have to determine the centre of mass of the system at a displacement of $s = 1{\text{m}}$ .
Now the x-component of the final velocity of the centre of mass will be the same as that of the initial velocity of the cracker i.e., ${v_x} = \dfrac{{10}}{{\sqrt 2 }}{\text{m}}{{\text{s}}^{ - 1}}$ .
->Step 2: Express the y-component of the final velocity using Newton’s third equation of motion to find the velocity of the centre of mass at $s = 1{\text{m}}$ .
The y-component of the final velocity of the centre of mass at the required height is given by Newton’s third equation of motion as ${v_y}^2 = {u_y}^2 - 2gs$ -------- (1)
Substituting for ${u_y} = \dfrac{{10}}{{\sqrt 2 }}{\text{m}}{{\text{s}}^{ - 1}}$ , $g = 10{\text{m}}{{\text{s}}^{ - 2}}$ and $s = 1{\text{m}}$ in equation (1) we get, ${v_y}^2 = {\left( {\dfrac{{10}}{{\sqrt 2 }}} \right)^2} - 2 \times 10 \times 1 = 30$
$ \Rightarrow {v_y} = \sqrt {30} {\text{m}}{{\text{s}}^{ - 1}}$
Then the velocity of the centre of mass at the given height is expressed as
${v_{CM}} = \sqrt {{v_x}^2 + {v_y}^2} $ -------- (2)
where ${v_x}$ and ${v_y}$ are the x-component and y-component of the velocity of the centre of mass.
Substituting for ${v_x} = \dfrac{{10}}{{\sqrt 2 }}{\text{m}}{{\text{s}}^{ - 1}}$ and ${v_y} = \sqrt {30} {\text{m}}{{\text{s}}^{ - 1}}$ in equation (2) we get, ${v_{CM}} = \sqrt {{{\left( {\dfrac{{10}}{{\sqrt 2 }}} \right)}^2} + {{\left( {\sqrt {30} } \right)}^2}} = \sqrt {80} = 4\sqrt 5 {\text{m}}{{\text{s}}^{ - 1}}$
Thus the required velocity of the centre of mass is ${v_{CM}} = 4\sqrt 5 {\text{m}}{{\text{s}}^{ - 1}}$ .
Hence the correct option is A.

Note:- In the projectile motion of the firecracker, there is no acceleration in the horizontal direction. So the velocity remains constant in that direction i.e., ${u_x} = {v_x}$ . Here the explosion is caused on its own. This implies that no external forces are present. The centre of mass of the cracker only changes if an external force acts on it. The acceleration due to gravity is taken to be negative in equation (1).