$A$ smooth sphere is moving on a horizontal surface with a velocity vector $(\,2\,\hat i + 2\,\hat j\,)$ $m/s$ immediately before it hit a vertical wall. The wall is parallel to vector $\hat j$ and coefficient of restitution between the sphere and the wall is $e = 1/2$ . The velocity of the sphere after it hits the wall is
$\,\hat i - \,\hat j$
$ - \,\hat i + 2\,\hat j$
$ - \,\hat i - \,\hat j$
$2\,\hat i - \,\hat j$
An unknown nucleus collides with a ${}^4He$ nucleus, and after the collision the two nuclei travel in perpendicular directions relative to each other. If kinetic energy is lost in the collision, the unknown nucleus must be
Two particles of masses ${m_1}$ and ${m_2}$ in projectile motion have velocities ${\vec v_1}$ and ${\vec v_2}$ respectively at time $t = 0$. They collide at time ${t_0}$. Their velocities become ${\vec v_1}'$ and ${\vec v_2}'$ at time $2{t_0}$ while still moving in air. The value of $|({m_1}\overrightarrow {{v_1}} '\, + {m_2}\overrightarrow {{v_2}} ') - ({m_1}\overrightarrow {{v_1}} \, + {m_2}\overrightarrow {{v_2}} )$| is
A body of mass $4\ kg$ collides head-on elastically with another body of mass $2\ kg$ kept at rest in free space. Time of collision is $0.02\ sec$ and average impulse force acted on each bodies is $100\ N$. Find the velocity of the $2\ kg$ body after the impact
The bob $A$ of a pendulum released from horizontal to the vertical hits another bob $B$ of the same mass at rest on a table as shown in figure.
If the length of the pendulum is $1\,m$, calculate
$(a)$ the height to which bob $A$ will rise after collision.
$(b)$ the speed with which bob $B$ starts moving.
Neglect the size of the bobs and assume the collision to be elastic.
A body falling on the ground from $a$ height of $10\, m$, rebounds to a height $2.5\, m,$ then the ratio of the velocities of the body just before and after the collision will be