A firecracker is thrown with a velocity of 30 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.The ra

Vedclass · Accepted Answer

$63 \, m$ or $117 \, m$

Vedclass · Answer

(D) The explosion force that splits the firecracker is internal to the system,so the path of the centre of mass of the system remains unchanged.The range of the centre of mass of the system is given by $R = \frac{u^2 \sin 2	heta}{g}$.Here,$u = 30 \, ms^{-1}$ and the angle with the horizontal is $	heta = 90^{\circ} - 75^{\circ} = 15^{\circ}$.Therefore,$R = \frac{30 	imes 30 	imes \sin(2 	imes 15^{\circ})}{10} = 90 	imes \sin(30^{\circ}) = 90 	imes 0.5 = 45 \, m$.So,the position ($x$-coordinate) of the centre of mass is at $45 \, m$ distance from the origin.Using the formula for the centre of mass,$X_{CM} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$,where $m_1 = m_2 = m$ and $X_{CM} = 45 \, m$:$45 = \frac{m(27) + m(x)}{m + m}$$45 = \frac{27 + x}{2}$$90 = 27 + x \Rightarrow x = 63 \, m$.However,the first piece could also fall at $-27 \, m$ (behind the origin) if the explosion is strong enough to reverse its direction:$45 = \frac{m(-27) + m(x)}{2m}$$90 = -27 + x \Rightarrow x = 117 \, m$.Thus,the other piece will fall at $63 \, m$ or $117 \, m$ from the shooting point.

Similar Questions

$A$ ball of mass $M$ moving with a speed of $2 \ m \ s^{-1}$ hits another ball of mass $1 \ kg$ moving in the same direction with a speed of $1 \ m \ s^{-1}$. If the kinetic energy of the center of mass is $4/3 \ J$,then the magnitude of $M$ is: (in $kg$)

$A$ body of mass $m_1 = 4 \text{ kg}$ moves at $5 \hat{i} \text{ m/s}$ and another body of mass $m_2 = 2 \text{ kg}$ moves at $10 \hat{i} \text{ m/s}$. The kinetic energy of the centre of mass is:

$A$ man of mass $M$ stands at one end of a plank of length $L$ which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is $3M$,the distance that the man moves relative to the ground is:

Are you in agreement with the statement: "The velocity of all particles of a system is equal to the velocity of the centre of mass"?

Two balls are thrown in the air. The second ball is thrown $2 \ s$ after the first ball. What is the acceleration of the center of mass of the two balls while they are in the air?

Vedclass Test Series

Exam Paper Generator

Online Exam Module