A body of mass $2\, kg$ moving with a velocity of $3\, m/sec$ collides head on with a body of mass $1\, kg$ moving in opposite direction with a velocity of $4\, m/sec$. After collision, two bodies stick together and move with a common velocity which in $m/sec$ is equal to
$1/4$
$1/3$
$2/3$
$3/4$
The force acting on a body moving along $x-$ axis varies with the position of the particle as shown in the figure. The body is in stable equilibrium at
A force acts on a $3\, gm$ particle in such a way that the position of the particle as a function of time is given by $x = 3t -4t^2 + t^3$ , where $x$ is in $meters$ and $t$ is in $seconds$ . The work done during the first $4\, second$ is ................. $\mathrm{mJ}$
Two identical $5\,\,kg.$ blocks are moving with same speed of $2\,\,m/s$ towards each other along a frictionless horizontal surface. The two blocks collide, stick together and come to rest. Consider to two blocks as a system, the work done on the system by the external forces will be .............. $\mathrm{Joule}$
Underline the correct alternative :
$(a)$ When a conservative force does positive work on a body, the potential energy of the body increases/decreases/remains unaltered.
$(b)$ Work done by a body against friction always results in a loss of its kinetic/potential energy.
$(c)$ The rate of change of total momentum of a many-particle system is proportional to the external force/sum of the internal forces on the system.
$(d)$ In an inelastic collision of two bodies, the quantities which do not change after the collision are the total kinetic energy/total linear momentum/total energy of the system of two bodies.
Work done in time $t $ on a body of mass $m$ which is accelerated from rest to a speed $v$ in time ${t_1}$ as a function of time $t$ is given by