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Multiple Choice Questions
Question 1.
Three bodies form an isolated system. There are m
_{1} = m
_{2} = 2m and m
_{3} = 3m. They have different direction, but all have the same initial speed v
_{0}. One or more elastic collision between the pair of the bodies where otherwise do not intersect. Find the maximum possible final speed of each of the three bodies.
(a) 3v
_{0}, 2v
_{0}, v
_{0}
(b) v
_{0}, 2v
_{0}, 3v
_{0}
(c) 2.4v
_{0}, 1.73v
_{0}, 1.41v
_{0}
(d) none of the above
Solution
Ans. (c)
Total initial kinetic energy = (1/2)m_{1}v_{0}^{2} + (1/2)m_{2}v_{0}^{2} + (1/3)m_{3}v_{0}^{2}=3mv_{0}^{2}
maximum speed will be realize Â if other particle speed are zero
So (1/2)m_{1}(v_{1max})^{2} = 3mv_{0}^{2}
Substituting for all m_{1}, m_{2}, m_{3}
we get
v_{1max} = 2.45v_{0}
v_{2max} = 1.73v_{0}
v_{3max} = 1.41v_{0}
Question 2.
A ball is dropped from height h on a floor where
Coefficient of restitution is e. Find the time required by the ball to stop rebounding
(a) âˆš(2h/g) (1 + e/1 - e)
(b) âˆš(2h/g) (1 + e)
(c) âˆš(2h/g) (1 - e/1 + e)
(d) none of the above
Solution
Ans. (a)
Time in first fall= âˆš(2h/g)
now velocity ofÂ ballÂ when it strike the floor
mgh = (1/2)mv^{2}
v = âˆš(2gh)
velocityÂ of rebound = eâˆš(2gh)
So time for the body to come again to floor= 2eâˆš(2gh)/g
So Time till now = âˆš(2h/g) + 2eâˆš(2h/g)
velocityÂ for the third rebound= e^{2}âˆš(2gh)
Time = 2e^{2}âˆš(2gh)/g
So Time till now = âˆš(2h/g) + 2eâˆš(2h/g)+ 2e^{2}âˆš(2gh)/g
So Total time
=âˆš(2h/g) + 2eâˆš(2h/g)+ 2e^{2}âˆš(2gh)/g + â€¦â€¦â€¦
$=\sqrt {\frac {2h}{g}}( 1 + 2 \sum_{1}^{\infty} e^n)$
$=\sqrt {\frac {2h}{g}} (\frac {1 + e}{1 - e})$
Â
Linked Type Comprehension
An object of mass M collides with a frictionless surface. The surface is assumed to be x-axis and object is coming at an angle Î¸ to the x-axis. The object bounces from the surface but the collision is not elastic
The initial velocity vector of the object
$v_i=(vcos{\theta})\mathbf{i}-(vsin{\theta})\mathbf{j}$
The vertical component of the velocity undergoes a change due to the impact. The magnitude is
fraction e of the original vertical components
The situation is depicted below in the figure
Question 3.
Find the velocity vector after the collision
(a) $v_f=(evcos{\theta})\mathbf{i}-(vsin{\theta})\mathbf{j}$
(b) $v_f=(vcos{\theta})\mathbf{i}+(evsin{\theta})\mathbf{j}$
(c) $v_f=(vcos{\theta})\mathbf{i}-(evsin{\theta})\mathbf{j}$
(d) $v_f=(evcos{\theta})\mathbf{i}+(vsin{\theta})\mathbf{j}$
Solution
Answer is (b)
Now
$v_f sin \theta _f=evsin \theta$ --(1)
$v_f cos \theta _f = vcos \theta$ --(2)
So final velocity vector
$v_f=(vcos{\theta})\mathbf{i}+(evsin{\theta})\mathbf{j}$
Question 4.
Find the final angle made with the horizontal
(a) $tan^{-1} (etan{\theta})$
(b) $cot^{-1}(etan{\theta})$
(c) $tan^{-1}(esin{\theta})$
(d) $tan^{-1}( ecos{\theta})$
Solution
Answer is (a)
From equation (1) and (2)
$tan \theta _f= \frac{evsin{\theta}}{vcos{\theta}}=etan{\theta}$
Or
$\theta_f=tan^{-1}( etan{\theta})$
Question 5.
Find the change in momentum $\Delta \mathbf{p}$ due to the impact
(a)$ mv(1+e)(cos{\theta})\mathbf{j}$
(b) $mv(1-e)(sin{\theta})\mathbf{j}$
(c) $mv(1-e)(cos{\theta})\mathbf{j}$
(d) $mv(1+e)(sin{\theta})\mathbf{j}$
Solution
(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
$\Delta \mathbf{p}=mv(1+e)(sin{\theta})\mathbf{j}$
Question 6.
Which one of the following is true about the case?
(a) Momentum remain conserved in the X â€“direction
(b) There is no loss of kinetic energy of the object
(c) The force during the collision acts in upward direction on the object
(d) The ratio of the final KE to the initial KE is $1-(1-e^2){sin}^2{\theta}$
Solution
(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
$K_f=\frac{1}{2}mv_f^2=\frac{mv^2}{2}\left[1-(1-e^2){sin}^2{\theta}\right]$
$K_i=\frac{1}{2}mv^2$
$\frac{K_f}{K_i}=1-(1-e^2){sin}^2{\theta}$
Multiple Choice Questions
Question 7.
Let
p_{1} and
p_{2} be the momentum of two equal particles before elastic collision and
p^{â€™}_{1} and
p_{2}^{â€™} be the momentum after collision.
Mow we know
p_{2} = 0
which of the following is true
(a) $\mathbf{p_1}.\mathbf{p_1}=\mathbf{p_1^\prime}.\mathbf{p_1^\prime}+\mathbf{p_2^\prime}.\mathbf{p_2^\prime}$
(b) $\mathbf{p_1}=\mathbf{p_1^\prime}+\mathbf{p_2^\prime}$
(c) $\mathbf{p_1^\prime}.\mathbf{p_2^\prime}=0$
(d) none of the above
Solution
(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
$\mathbf{p_1^\prime}.\mathbf{p_2^\prime}=0$
Question 8.
During inelastic collision between two bodies, which of the following quantities always remain conserved?
(a) Total kinetic energy.
(b) Total mechanical energy.
(c) Total linear momentum.
(d) Speed of each body
Solution
(c). Total linear momentum remains conserved
Question 9. In a collinear collision, a particle with an initial speed $v_0$ strikes a stationary particle of the same mass. If the final total kinetic energy is 50% greater than the original kinetic energy, the magnitude of the relative velocity between the two particles, after collision, is
(a) $\sqrt {2} v_0$
(b) $\frac {v_0}{2}$
(c) $\frac {v_0}{\sqrt {2}}$
(d) $\frac {v_0}{4}$
Solution
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
$|v_1 - v_2| =\sqrt {2} v_0$
Question 10.
Statement - I : A point particle of mass m moving with speed v collides with stationary point particle of mass M. If the maximum energy loss possible is given as $f(\frac {1}{2} mv^2)$ then $f=\frac {m}{M+m}$
Statement 0 II : Maximum energy loss occurs when the particles get stuck together as a result of the collision.
(a) Statement-I is true, Statement-II is true, Statement-II is a correct explanation of Statement-I.
(b) Statement-I is true, Statement-II is true, Statement-II is a not correct explanation ofStatement-I.
(c) Statement-I is true, Statement-II is false.
(d) Statement-I is false, Statement-II is true
Solution
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
Question 11.
A set of n identical cubical blocks lies at rest parallel to each other along a line on a smooth horizontal surface. The separation between the near surfaces of any two adjacent blocks is L. The block at one end is given a speed v towards the next one at time t=0. All collisions are completely inelastic, then
(a) The last block starts moving at $t=\frac {(n-1)L}{v}$
(b) The last block starts moving at $t=\frac {n(n-1)L}{2v}$
(c) The centre of mass of the system will have a final speed v
(d) The centre of mass of the system will have a final speed $\frac {v}{n}$
Solution
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
Question 12.
A body of mass m moving with velocity v collides head on with another body of mass 2m which is initially at rest. The ratio of K.E. of colliding body before and after collision will be
(a) 1 : 1
(b) 2 : 1
(c) 4 : 1
(d) 9 : 1
Solution
(d) is the answer
Subjective Numericals
Question 13
A body A having mass m
_{1} travellingÂ with velocityÂ u Â makes an head on elastic collision with the stationary bodyÂ B of mass m
_{2}
(a) Calculate the final velocities of A and B
(b) Calculate the Final Kinetic energy of both the mass
(c) Calculate the ratio of the kinetic energy transferred to m
_{2} to the original kinetic energy
(d) For what value of m
_{2}, all the energy is transferred to the stationary object
(e) Calculate the
velocity of the Center of mass before collision and after collision
Solution
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}$
Question 14.
An imperfectly elastic particle is projected from a point in horizontal plane with velocity u at any angle Î± to the horizon. If e is the
Coefficient of restitution
Let
i and
j are the unit vector across the x and y axis respectively
(a) Find the velocity of particle after first rebound
(b) Find the total time taken by the particle before stopping rebounding
(c) Find the total range
(d) Find the velocity at the m
^{th} rebound
(e)Find the tangent ofÂ angle of projection at m
^{th} rebound
(g) Find the height reached after m
^{th} rebound
(f) Find the total impulse exerted by the surface on the ball
Solution
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
$=musin \alpha[ 1 + 2(e +e^2+ e^3 +.......)]$
$=musin{\alpha}\left(\frac{1+e}{1-e}\right)$
Question 15.
A body of mass 3 kg moving with a velocity (
i +2
j +3
k) m/s collides with another body of mass 4 kg moving with a velocity (2
i+
j+
k) in m/s.They stick togetherÂ .Find the velocity of the composite body.Find the Kinetic energy before collision and after collision
Solution
Law of conservation of momentum
mv_{1}+ Mv_{2}= (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
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