A free body of mass 8 , kg is travelling at 2 | vedclass

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

Question

(B) As the body splits into two equal parts due to an internal explosion,the momentum of the system remains conserved.Initial momentum = Final momentu

Vedclass · Accepted Answer

One part comes to rest and the other moves in the same direction as that of the original body.

Vedclass · Answer

(B) As the body splits into two equal parts due to an internal explosion,the momentum of the system remains conserved.Initial momentum = Final momentum$8 	imes 2 = 4v_1 + 4v_2$$\Rightarrow v_1 + v_2 = 4$ ...$(i)$By the law of conservation of energy:Initial kinetic energy + Energy released = Final kinetic energy$\frac{1}{2} 	imes 8 	imes (2)^2 + 16 = \frac{1}{2} 	imes 4 	imes v_1^2 + \frac{1}{2} 	imes 4 	imes v_2^2$$16 + 16 = 2v_1^2 + 2v_2^2$$v_1^2 + v_2^2 = 16$ ...(ii)From $(i)$,$v_2 = 4 - v_1$. Substituting this into (ii):$v_1^2 + (4 - v_1)^2 = 16$$v_1^2 + 16 - 8v_1 + v_1^2 = 16$$2v_1^2 - 8v_1 = 0$$2v_1(v_1 - 4) = 0$So,$v_1 = 0$ or $v_1 = 4 \, m/s$.If $v_1 = 0$,then $v_2 = 4 \, m/s$. If $v_1 = 4$,then $v_2 = 0$.Thus,one part comes to rest and the other moves in the same direction as that of the original body.

$A$ free body of mass $8 \, kg$ is travelling at $2 \, m/s$ in a straight line. At a certain instant,the body splits into two equal parts due to an internal explosion which releases $16 \, J$ of energy. Neither part leaves the original line of motion. Finally:

Similar Questions

$A$ wooden block of mass $M$ is suspended by a string. $A$ bullet of mass $m$ passes through the block with velocity $v$ and emerges with a velocity $v/2$ in the same direction. To what height will the block rise?

Block $A$ of mass $m$ is hanging from a vertical spring of spring constant $k$ and is at rest. Block $B$ of mass $m$ strikes block $A$ with velocity $v$ and sticks to it. The velocity $v$ for which the spring just attains its natural length is:

$A$ sphere of mass $m$ moving with a constant velocity $u$ hits another stationary sphere of the same mass. If $e$ is the coefficient of restitution,then the ratio of the velocity of the two spheres after the collision will be

$A$ particle is moving in a circle of radius $r$ under the action of a force $F = \alpha r^2$ which is directed towards the centre of the circle. The total mechanical energy (kinetic energy + potential energy) of the particle is (take potential energy $= 0$ for $r = 0$).

$A$ block moving horizontally on a smooth surface with a speed of $40\, m/s$ splits into two parts with masses in the ratio of $1:2$. If the smaller part moves at $60\, m/s$ in the same direction,then the fractional change in kinetic energy is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module