Two particles of masses $m_1, m_2$ move with initial velocities $u_1$and $u_2$ On collision, one of the particles get excited to higher level, after absorbing energy $\varepsilon $. If final velocities of particles be $v_1$ and $v_2$ then we must have
$\frac{1}{2}{m_1}{u_1}^2 + \frac{1}{2}{m_2}{u_2}^2 = \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2 - \varepsilon $
$\;\frac{1}{2}{m_1}{u_1}^2 + \frac{1}{2}{m_2}{u_2}^2 - \varepsilon = \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2$
$\;\frac{1}{2}{m_1}{u_1}^2 + \frac{1}{2}{m_2}{u_2}^2 + \varepsilon = \frac{1}{2}{m_1}{v_1}^2 + \frac{1}{2}{m_2}{v_2}^2$
$m_1^2u_1+m_2^2u_2 - \varepsilon = m_1^2v_1+m_2^2v_2$
A ball is allowed to fall from a height of $10 \,m$. If there is $40 \%$ loss of energy due to impact, then after one impact ball will go up by ........ $m$
An isolated rail car of mass $M$ is moving along a straight, frictionless track at an initial speed $v_0$. The car is passing under a bridge when $a$ crate filled with $N$ bowling balls, each of mass $m$, is dropped from the bridge into the bed of the rail car. The crate splits open and the bowling balls bounce around inside the rail car, but none of them fall out. Is the momentum of the rail car $+$ bowling balls system conserved in this collision?
A particle is moving along a vertical circle of radius $R$. At $P$, what will be the velocity of particle (assume critical condition at $C)$ ?
Separation of Motion of a system of particles into motion of the centre of mass and motion about the centre of mass
$(a)$ Show $p = p _{t}^{\prime}+m_{t} V$
where $p$, is the momentum of the the particle (of mass $m$ ) and $p_{t}^{\prime \prime}=m_{t} v_{t}$,
Note $v_{t}$, is the velocity of the particle relative to the centre of mass. Also, prove using the definition of the centre of mass $\sum p _{t}^{\prime}=0$
$(b)$ Show $K=K^{\prime}+1 / 2 M V^{2}$
where $K$ is the total kinetic energy of the system of particles. $K^{\prime}$ is the total kinetic energy of the system when the particle velocities are taken with respect to the centre of mass and $M V^{2} / 2$ is the kinetic energy of the translation of the system as a whole (i.e. of the centre of mass motion of the system).
$(c)$ Show $L = L ^{\prime}+ R \times M V$
where $L ^{\prime}=\sum r _{t}^{\prime} \times p _{t}^{\prime}$ is the angular momentum of the system about the centre of mass with velocities taken relative to the centre of mass. Remember $r _{t}^{\prime}= r _{t}- R$; rest of the notation is the standard notation used in the chapter. Note $L$ ' and $M R \times V$ can be said to be angular momenta, respectively, about and of the centre of mass of the system of particles.
$(d)$ Show $\frac{d L ^{\prime}}{d t}=\sum r _{t}^{\prime} \times \frac{d p ^{\prime}}{d t}$
Further, show that
$\frac{d L ^{\prime}}{d t}=\tau_{e x t}^{\prime}$
where $\tau_{c t t}^{\prime}$ is the sum of all external torques acting on the system about the centre of mass. (Hint: Use the definition of centre of mass and third law of motion. Assume the internal forces between any two particles act along the line joining the particles.)
A man is standing on a cart of mass double the mass of man. Initially cart is at rest. Now man jumps horizontally with relative velocity $'u'$ with respect to cart. Then work done by internal forces of the man during the process of jumping will be :