$A$ body of mass $5\,kg$ is moving with a momentum of $10\,kg\,m/s$. Now a force of $2\,N$ acts on the body in the direction of its motion for $5\,s$. The increase in the kinetic energy of the body is $...........J$.

  • A
    $30$
  • B
    $29$
  • C
    $28$
  • D
    $27$

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Similar Questions

$A$ 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:

$A$ pendulum of length $2 \; m$ consists of a wooden bob of mass $50 \; g$. $A$ bullet of mass $75 \; g$ is fired towards the stationary bob with a speed $v$. The bullet emerges out of the bob with a speed $\frac{v}{3}$ and the bob just completes the vertical circle. The value of $v$ is $\dots \; ms^{-1}$. (if $g = 10 \; m/s^2$)

Fill in the blanks:
$(a)$ When an object is lifted from the ground to a certain height,the work done against the gravitational force is ......
$(b)$ When the work done is zero,the speed of the object remains ..........
$(c)$ For a .......... collision,the coefficient of restitution is $1$.

$A$ particle of mass $m$ moving horizontally with velocity $v_0$ strikes a smooth wedge of mass $M$,as shown in the figure. After the collision,the particle starts moving up the inclined face of the wedge and rises to a height $h$. Choose the correct statement$(s)$ related to particle $m$.

In two separate collisions,the coefficients of restitution $e_1$ and $e_2$ are in the ratio $3:1$. In the first collision,the relative velocity of approach is twice the relative velocity of separation. Find the ratio between the relative velocity of approach and the relative velocity of separation in the second collision.

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