Elastic and Inelastic Collisions problems and solutions

Ans. (c)
Total initial kinetic energy = (1/2)m₁v₀² + (1/2)m₂v₀² + (1/3)m₃v₀²=3mv₀²
maximum speed will be realize  if other particle speed are zero
So (1/2)m₁(v_1max)² = 3mv₀²
Substituting for all m₁, m₂, m₃
we get
v_1max = 2.45v₀
v_2max = 1.73v₀
v_3max = 1.41v₀

Ans. (a)
Time in first fall= √(2h/g)
now velocity of  ball  when it strike the floor
mgh = (1/2)mv²
v = √(2gh)
velocity  of rebound = e√(2gh)
So time for the body to come again to floor= 2e√(2gh)/g
So Time till now = √(2h/g) + 2e√(2h/g)
velocity  for the third rebound= e²√(2gh)
Time = 2e²√(2gh)/g
So Time till now = √(2h/g) + 2e√(2h/g)+ 2e²√(2gh)/g
So Total time
=√(2h/g) + 2e√(2h/g)+ 2e²√(2gh)/g + ………
$=\sqrt {\frac {2h}{g}}( 1 + 2 \sum_{1}^{\infty} e^n)$
$=\sqrt {\frac {2h}{g}} (\frac {1 + e}{1 - e})$
 

Answer is (b)
Now
$v_f sin \theta _f=evsin \theta$ --(1)
$v_f cos \theta _f = vcos \theta$ --(2)

So final velocity vector
$v_f=(vcos{\theta})\mathbf{i}+(evsin{\theta})\mathbf{j}$

Answer is (a)
From equation (1) and (2)
$tan \theta _f= \frac{evsin{\theta}}{vcos{\theta}}=etan{\theta}$
Or
$\theta_f=tan^{-1}( etan{\theta})$

(d) is the answer
$v_i=(vcos{\theta})\mathbf{i}-(vsin{\theta})\mathbf{j}$
$v_f=(vcos{\theta})\mathbf{i}+(evsin{\theta})\mathbf{j}$

$\Delta \mathbf{p}=m\mathbf{v_f} - m\mathbf{v_i}$
Substituting the values
$\Delta \mathbf{p}=mv(1+e)(sin{\theta})\mathbf{j}$

(a) ,(c) ,(d) are correct
Since frictional force are null, horizontal components of the momentum remain conserved.

The force acts in upward direction, as change in momentum is in upward direction

$\mathbf{F}\Delta t=\Delta \mathbf{p}$
$\mathbf{F}=\frac{\Delta \mathbf{p}}{\Delta t}$
As $\Delta \mathbf{p}$ -> $\mathbf{j}$ direction
$\Delta \mathbf{F}$ -> $\mathbf{j}$ direction

$K_f=\frac{1}{2}mv_f^2=\frac{mv^2}{2}\left[1-(1-e^2){sin}^2{\theta}\right]$
$K_i=\frac{1}{2}mv^2$
$\frac{K_f}{K_i}=1-(1-e^2){sin}^2{\theta}$

(a), (b), (c)
(a) is the law of conservation of energy
$\mathbf{p_1}.\mathbf{p_1}=\mathbf{p_1^\prime}.\mathbf{p_1^\prime}+\mathbf{p_2^\prime}.\mathbf{p_2^\prime}$
(b) is law of conservation of linear momentum

$\mathbf{p_1}=\mathbf{p_1^\prime}+\mathbf{p_2^\prime}$ ---(2)
Squaring equation (2)
$p_1^2={p'}_1^2+{p'}_2^2+2\mathbf{p_1^\prime}.\mathbf{p_2^\prime}$
Or
$\mathbf{p_1^\prime}.\mathbf{p_2^\prime}=0$

(c). Total linear momentum remains conserved

Answer is (a)
From conservation of momentum
$mv_0= mv_1 + mv_2$
or
$v_0=v_1 + v_2$ -(1)
Given from questions
$\frac {1}{2}mv_1^2 + \frac {1}{2}mv_2^2= \frac {3}{2} \frac {1}{2}mv_0^2$
$v_1^2 + v_2^2 = \frac {3}{2}v_0^2$ -(2)
Squaring (1)
$v_0^2 = v_1^2 + v_2^2 + 2v_1 v_2$
or
$2v_1 v_2 = -\frac {v_0^2}{2}$
Now
$(v_1- v_2)^2 = v_1^2 + v_2^2 -2 v_1v_2 = 2v_0^2$
or
$|v_1 - v_2| =\sqrt {2} v_0$

Maximum energy loss happens when particles stick together
Then
mv = (M+m)V
or
$V= \frac {mv}{M+m}$
Now
Loss of KE = $\frac {1}{2} (M+m)V^2 - \frac {1}{2} mv^2$
Substituting the values of V from above
Loss of KE=$\frac {M}{m+M} \frac {1}{2} mv^2$
Hence (d) is correct option

Since collision is perfectly inelastic so all the blocks will stick together one by one and move in a form of combined mass.
Time required to cover a distance L by first block =L/v
Now first and second block will stick together and move with v/2 velocity (by applying conservation of momentum) and combined system will take time L/v/2=2L/v to reach up to block third.
Now these three blocks will move with velocity v/3 and combined system will take time L/v/3=3L/v to reach upto the block fourth.
So, total time =$\frac {L}{v}+ \frac {2L}{v}+ \frac {3L}{v} + ... + \frac {(n-1)L}{v}=\frac {n(n-1)L}{2v}$
Velocity of combined system having n blocks as v/n.
Hence (b) is the answer

(d) is the answer

Let $v_1$ and $v_2$ be the velocities after the collision
Law of conservation of linear momentum
$m_1u=m_1v_1+m_2v_2$ ---(1)
For elastic collision
Velocity of approach=-velocity of separation
$u=-(v_1-v_2)$
or
$v_2-v_1=u$ ---(2)

Solving equation (1) and (2)
We get
$v_1=\left(\frac{m_1-m_2}{m_1+m_2}\right)u$

$v_2=\left(\frac{2m_1}{m_1+m_2}\right)u$

Initial Kinetic energy ($K_{1i}$ )
$=\frac{1}{2}m_1u^2$

KE of the mass $m_1$( $K_{1f}$)
$=\frac{1}{2}m_1v_1^2$
$=\frac{1}{2}m_1\left(\frac{m_1-m_2}{m_1+m_2}\right)^2u^2$
KE of mass $m_2$($K_{2f}$ )
$=\frac{1}{2}m_2v_2^2$
$=\frac{1}{2}m_2\left(\frac{2m_1}{m_1+m_2}\right)^2u^2$
Now
$\frac{K_{2f}}{K_{1i}}=\frac{\frac{1}{2}m_2\left(\frac{2m_1}{m_1+m_2}\right)^2u^2}{\frac{1}{2}m_1u^2}$
$=\frac{4m_1m_2}{(m_1+m_2)^2}$
$=1-\frac{(m_1-m_2)^2}{(m{_1^}+m_2)^2}$

Now when $m_1=m_2$
$K_{2f}=K_{1i}$
Velocity of center of mass before collision
$=\frac{m_1u}{m_1+m_2}$
Since no external force is acting, Center of mass velocity will not change
Velocity of center of mass after collision
$=\frac{m_1u}{m_1+m_2}$

Now we know from projectile motion
Velocity of strike =$(ucos{\alpha})\mathbf{i}-(usin{\alpha})\mathbf{j}$
Since no force acts in the horizontal direction, Horizontal velocity will be same be same after
Vertical velocity will experience the force
So vertical velocity after rebound =$eu sin \alpha \mathbf{j}$

So velocity of particle after first rebound
$=(ucos{\alpha})\mathbf{i}+(eusin{\alpha})\mathbf{j}$

Now in case of Projectile motion
Time of flight =$\frac{2usin{\alpha}}{g}=\frac{2u_y}{g}$
So time before striking first
$T_1 =\frac{2usin{\alpha}}{g}$
Time taken after first strike and before second strike
$T_2=\frac{2eusin{\alpha}}{g}$
Now velocity of particle after second strike is given by
$=(ucos{\alpha})\mathbf{i}+(e^2usin{\alpha})\mathbf{j}$

Time taken after second strike and before third strike
$T_3=\frac{2e^2usin{\alpha}}{g}$
Similarly other can find out
So total time
$=T_1+T_2+T_3+ ...$
$=\frac{2usin{\alpha}}{g}\left(1+e+e^2+................\right)$
$=\frac{2usin{\alpha}}{g}\left(\frac{1}{1-e}\right)$

Now range is given by
$R=u_xT_1+u_xT_2 + ....$
$=u_xT$
$=ucos{\alpha}\left[\frac{2usin{\alpha}}{g}\left(\frac{1}{1-e}\right)\right]$
$=\frac{u^2sin{2}\alpha}{g(1-e)}$

Velocity after mth rebound
$=(ucos{\alpha})\mathbf{i}+(e^musin{\alpha})\mathbf{j}$
Tangent of angle of projection after mth rebound
$=\frac{e^m u sin{\alpha}}{ucos{\alpha}}$
$=e^m tan \alpha$

We know from projectile
$\text {Height}=\frac{u^2{sin}^2{\alpha}}{2g}$

So height after m rebound
$=\frac{(e^musin{\alpha})^2}{2g}$

Impluse exerted by the surface on the ball
$I_1= m[eu sin \alpha \mathbf{j} (-usin \alpha \mathbf{j})]$
$=mu sin \alpha ( 1+e) \mathbf{j}$

Similarly
$I_2= musin \alpha( e+e^2) \mathbf{j}$

$I_3= musin \alpha ( e^2+e^3) \mathbf{j}$
So total impulse
$=musin \alpha[ 1 + 2(e +e^2+ e^3 +.......)]$
$=musin{\alpha}\left(\frac{1+e}{1-e}\right)$

Law of conservation of momentum
mv₁+ Mv₂= (m+M) v

3(i +2j +3k) + 4(2i+j+k) =7v
7v=11i+10j+13k
v=(11/7)I + (10/7)j +(13/7)k

Magnitude of velocity of mass 3 kg=(1+4+9)^.5
KE =(1/2)*3*14=21 J

Magnitude of velocity of mass 4 kg=(4+1+1)^.5
KE =(1/2)*4*6=12 J

Magnitude of velocity of composite body
=(121+100+169)^.5 *(1/7)

KE =(1/2)*7*(121+100+169) *(1/7)=195/7=27.8 J