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Center of Mass and Collision

JEE Physics · 64 questions · Page 3 of 7 · Click an option or "Show Solution" to reveal answer

Q21
Consider a two particle system with particles having masses m1{m_1} and m2{m_2}. If the first particle is pushed towards the center of mass through a distance d,d, by what distance should the second particle is moved, so as to keep the center of mass at the same position?
A m2m1  d{{{m_2}} \over {{m_1}}}\,\,d
B m1m1+m2d{{{m_1}} \over {{m_1} + {m_2}}}d
C m1m2d{{{m_1}} \over {{m_2}}}d
D dd
Correct Answer
Option C
Solution

Initially,

0=m1(−x1)+m2x2m1+m2⇒m1x1=m2x20 = {{{m_1}\left( { - {x_1}} \right) + {m_2}{x_2}} \over {{m_1} + {m_2}}} \Rightarrow {m_1}{x_1} = {m_2}{x_2}

Finally, mass m1 moved towards the center a distance d so the distance of mass m1 from the origin is x1 - d, and now let mass m2 need to move d' to keep the center at the origin.

∴\therefore

0=m1(d−x1)+m2(x2−d′)m1+m20 = {{{m_1}\left( {d - {x_1}} \right) + {m_2}\left( {{x_2} - d'} \right)} \over {{m_1} + {m_2}}}
⇒0=m1d−m1x1+m2x2−m2d′\Rightarrow 0 = {m_1}d - {m_1}{x_1} + {m_2}{x_2} - {m_2}d'
⇒d′=m1m2d\Rightarrow d' = {{{m_1}} \over {{m_2}}}d
Q22
A proton of mass m collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90o with respect to each other. The mass of unknown particle is :
A m2{m \over 2}
B m
C m3{m \over {\sqrt 3 }}
D 2 m
Correct Answer
Option B
Solution

In figure (i) before collision, m' is mass of unknown particle; m is mass of proton; v1 is initial velocity. (i) Before collision : Now, in figure (ii), v1 is final velocity of unknown particle and v2 is final velocity of proton. (ii) After collision : By conservation of momentum, we have Momentum before collision = Momentum after collision Consider x-component, we have

mvi+m′ . 0=m′v1cos⁡45∘+mv2cos⁡45∘m{v_i} + m'\,.\,0 = m'{v_1}\cos 45^\circ + m{v_2}\cos 45^\circ
mvi=12(m′v1+mv2)m{v_i} = {1 \over {\sqrt 2 }}(m'{v_1} + m{v_2})

..... (1) Consider y-component, we have

0=m′v1sin⁡45∘−mv2sin⁡45∘0 = m'{v_1}\sin 45^\circ - m{v_2}\sin 45^\circ
12(m′vi−mv2)=0⇒m′v1=mv2{1 \over {\sqrt 2 }}(m'{v_i} - m{v_2}) = 0 \Rightarrow m'{v_1} = m{v_2}

...... (2) Substitute Eq. (2) in Eq. (1), we get

mvi=12(mv2+mv2)−2mv2m{v_i} = {1 \over {\sqrt 2 }}(m{v_2} + m{v_2}) - \sqrt 2 m{v_2}
⇒vi=2v2\Rightarrow {v_i} = \sqrt 2 {v_2}

..... (3) Using Eq. (2) and (3) in Eq. (1), we get

m2v2=12(m′v1+m′v1)m\sqrt 2 {v_2} = {1 \over {\sqrt 2 }}(m'{v_1} + m'{v_1})
2mv2=2m′v12m{v_2} = 2m'{v_1}

(∵\because

v1=v2{v_1} = {v_2}

)

⇒m=m′\Rightarrow m = m'
Q23
A particle of mass mm moving in the xx direction with speed 2v2v is hit by another particle of mass 2m2m moving in the yy direction with speed v.v. If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:
A 56%56\%
B 62%62\%
C 44%44\%
D 50%50\%
Correct Answer
Option A
Solution

Applying conservation of linear momentum in x direction

2mv=(2m+m)Vx2mv = \left( {2m + m} \right){V_x}
⇒Vx=23v\Rightarrow {V_x} = {2 \over 3}v

Applying conservation of linear momentum in y direction

2mv=(2m+m)Vy2mv = \left( {2m + m} \right){V_y}
⇒Vy=23v\Rightarrow {V_y} = {2 \over 3}v

Final speed of 3m mass,

Vf{V_f}

=

Vx2+Vy2\sqrt {V_x^2 + V_y^2}

=

4v29+4v29\sqrt {{{4{v^2}} \over 9} + {{4{v^2}} \over 9}}

=

8v29\sqrt {{{8{v^2}} \over 9}}

Initial kinetic energy

Ei=12m(2v)2+12(2m)(v)2{E_i} = {1 \over 2}m{\left( {2v} \right)^2} + {1 \over 2}\left( {2m} \right){\left( v \right)^2}
=2mv2+mv2= 2m{v^2} + m{v^2}

=

3mv23m{v^2}

Final kinetic energy,

Ef=12(3m){E_f} = {1 \over 2}\left( {3m} \right)
Vf2{V_f^2}
=3m2[8v29]= {{3m} \over 2}\left[ {{{8{v^2}} \over 9}} \right]
=4mv23= {{4m{v^2}} \over 3}

Energy loss =

Ei−Ef{E_i} - {E_f}

=

3mv2−4mv233m{v^2} - {{4m{v^2}} \over 3}

=

5mv23{{5m{v^2}} \over 3}

Percentage loss in the energy during the collision =

Ei−EfEi{{{E_i} - {E_f}} \over {{E_i}}}
×100\times 100

=

5mv233mv2{{{{5m{v^2}} \over 3}} \over {3m{v^2}}}
×100\times 100

=

59×100{5 \over 9} \times 100
≃56%\simeq 56\%
Q24
A body of mass 2 kg makes an eleastic collision with a second body at rest and continues to move in the original direction but with one fourth of its original speed. What is the mass of the second body ?
A 1.2 kg
B 1.0 kg
C 1.8 kg
D 1.5 kg
Correct Answer
Option A
Solution

By conservation of linear momentum:

2v0=2(v04)+mv⇒2v0=v02+mv2{v_0} = 2\left( {{{{v_0}} \over 4}} \right) + mv \Rightarrow 2{v_0} = {{{v_0}} \over 2} + mv
⇒3v02=mv  ...(1)\Rightarrow {{3{v_0}} \over 2} = mv\,\,...(1)

Since collision is elastic

Vseparation=Vapproch{V_{separation}} = {V_{approch}}
⇒v−v04=v0⇒m=65=1.2 kg\Rightarrow v - {{{v_0}} \over 4} = {v_0} \Rightarrow m = {6 \over 5} = 1.2\,kg
Q25
A simple pendulum, made of a string of length ℓ\ell and a bob of mass m, is released from a small angle θ0{{\theta _0}}. It strikes a block of mass M, kept on a horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle θ1{{\theta _1}}. Then M is given by :
A m2(θ0+θ1θ0−θ1){m \over 2}\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)
B m2(θ0−θ1θ0+θ1){m \over 2}\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)
C m(θ0+θ1θ0−θ1)m\left( {{{{\theta _0} + {\theta _1}} \over {{\theta _0} - {\theta _1}}}} \right)
D m(θ0−θ1θ0+θ1)m\left( {{{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}}} \right)
Correct Answer
Option C
Solution

v =

2gℓ(1−cos⁡θ0)\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)}

v1 =

2gℓ(1−cos⁡θ1)\sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)}

By momentum conservation m

2gl(1−cos⁡θ0)\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}
=MVm−m2g(1−cos⁡θ)= M{V_m} - m\sqrt {2g\left( {1 - \cos \theta } \right)}

⇒\Rightarrow

m2gℓ{1−cos⁡θ0+1−cos⁡θ1}m\sqrt {2g\ell } \left\{ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right\}

== MVm and e = 1 =

Vm+2gℓ(1−cos⁡θ1)2gℓ(1−cos⁡θ0){{{V_m} + \sqrt {2g\ell \left( {1 - \cos {\theta _1}} \right)} } \over {\sqrt {2g\ell \left( {1 - \cos {\theta _0}} \right)} }}
2gℓ\sqrt {2g\ell }
(1−cos⁡θ0−1−cos⁡θ1)\left( {\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} } \right)

== Vm . . .(I) m

2gℓ(1−cos⁡θ0+1−cos⁡θ1)\sqrt {2g\ell } \left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)

== MVM . . .(II) Dividing

(1−cos⁡θ0+1−cos⁡θ1)(1−cos⁡θ0+1−cos⁡θ1)=Mm{{\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)} \over {\left( {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right)}} = {M \over m}

By componendo divided

m−Mm+M{{m - M} \over {m + M}}

=

1−cos⁡θ11−cos⁡θ0=sin⁡(θ12)sin⁡(θ02){{\sqrt {1 - \cos {\theta _1}} } \over {\sqrt {1 - \cos {\theta _0}} }} = {{\sin \left( {{{{\theta _1}} \over 2}} \right)} \over {\sin \left( {{{{\theta _0}} \over 2}} \right)}}

⇒\Rightarrow

Mm=θ0−θ1θ0+θ1⇒M=θ0−θ1θ0+θ1{M \over m} = {{{\theta _0} - {\theta _1}} \over {{\theta _0} + {\theta _1}}} \Rightarrow M = {{{\theta _0} - \theta 1} \over {{\theta _0} + {\theta _1}}}
Q26
A body of mass M at rest explodes into three pieces, in the ratio of masses 1 : 1 : 2. Two smaller pieces fly off perpendicular to each other with velocities of 30 ms−-1 and 40 ms−-1 respectively. The velocity of the third piece will be :
A 15 ms−-1
B 25 ms−-1
C 35 ms−-1
D 50 ms−-1
Correct Answer
Option B
Solution

Given problem a body of mass MM explodes into three pieces of mass ratio 1:1:21: 1: 2 ∴\therefore Mass of fragments will be x,x,2xx, x, 2 x Hence, M=x+x+2x=4x kgM=x+x+2 x=4 x \mathrm{~kg} As in the process of explosion no external forces are involved and explosion occurs due to internal forces.

Thus, momentum of the system will be conserved.

pinitial =pfinal p_{\text{initial }}=p_{\text{final }} By law of conservation of momentum,

M×0=M4×30i^+M4×40j^+2M4v⃗M \times 0=\frac{M}{4} \times 30 \hat{i}+\frac{M}{4} \times 40 \hat{j}+\frac{2 M}{4} \vec{v}

Where v⃗\vec{v} is the velocity of the third fragment.

Mv⃗2=−M4(30i^+40j^)\frac{M \vec{v}}{2} =-\frac{M}{4}(30 \hat{i}+40 \hat{j})

⇒\Rightarrow

v⃗=−15i^−20j^\vec{v} =-15 \hat{i}-20 \hat{j}

Thus, magnitude of v⃗=∣v⃗∣=vx2+vy2\vec{v}=|\vec{v}|=\sqrt{v_x^2+v_y^2} =(−15)2+(−20)2=\sqrt{(-15)^2+(-20)^2}

∣v⃗∣=625=25 m/s|\vec{v}|=\sqrt{625}=25 \mathrm{~m} / \mathrm{s}
Q27
A ball of mass 200 g200 \mathrm{~g} rests on a vertical post of height 20 m20 \mathrm{~m}. A bullet of mass 10 g10 \mathrm{~g}, travelling in horizontal direction, hits the centre of the ball. After collision both travels independently. The ball hits the ground at a distance 30 m30 \mathrm{~m} and the bullet at a distance of 120 m120 \mathrm{~m} from the foot of the post. The value of initial velocity of the bullet will be (if g=10 m/s2g=10 \mathrm{~m} / \mathrm{s}^{2}) :
A 120 m/s
B 360 m/s
C 400 m/s
D 60 m/s
Correct Answer
Option B
Solution

∵\because Time of flight of each ball and bullet

=2Hg=2×2010=2= \sqrt {{{2H} \over g}} = \sqrt {{{2 \times 20} \over {10}}} = 2

s ⇒\Rightarrow By applying linear momentum conservation

100u+200(0)=200(302)+10(1202)100u + 200(0) = 200\left( {{{30} \over 2}} \right) + 10\left( {{{120} \over 2}} \right)
u=360u = 360

m/s

Q28
A player caught a cricket ball of mass 150150 gg moving at a rate of 2020 m/s.m/s. If the catching process is completed in 0.1s,0.1s, the force of the blow exerted by the ball on the hand of the player is equal to
A 150150 NN
B 33 NN
C 3030 NN
D 300300 NN
Correct Answer
Option C
Solution

We know, Force×\times time = Impulse = Change in momentum ∴\therefore

F×t=m(v−u)F \times t = m\left( {v - u} \right)

⇒\Rightarrow

F=m(v−u)t=0.15(0−20)0.1=30NF = {{m\left( {v - u} \right)} \over t} = {{0.15\left( {0 - 20} \right)} \over {0.1}} = 30N
Q29
A particle of mass m is moving in a straight line with momentum p. Starting at time t = 0, a force F = kt acts in the same direction on the moving particle during time interval T so that its momentum changes from p to 3p. Here k is a constant. The value of T is :
A 2kp2\sqrt {{k \over p}}
B 2pk2\sqrt {{p \over k}}
C 2p2\sqrt {{{2p} \over 2}}
D 2kp\sqrt {{{2k} \over p}}
Correct Answer
Option B
Solution
dpdt=F=kt{{dp} \over {dt}} = F = kt
∫P3PdP=∫0Tkt dt\int_P^{3P} {dP} = \int_0^T {kt\,dt}
2p=KT222p = {{K{T^2}} \over 2}
T=2PKT = 2\sqrt {{P \over K}}
Q30
Two bodies of mass 1 kg1 \mathrm{~kg} and 3 kg3 \mathrm{~kg} have position vectors i^+2j^+k^\hat{i}+2 \hat{j}+\hat{k} and −3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k} respectively. The magnitude of position vector of centre of mass of this system will be similar to the magnitude of vector :
A i^+2j^+k^\hat{i}+2 \hat{j}+\hat{k}
B −3i^−2j^+k^-3 \hat{i}-2 \hat{j}+\hat{k}
C −2i^+2k^-2 \hat{i}+2 \hat{k}
D 2i^−j^+2k^2 \hat{i}-\hat{j}+2 \hat{k}
Correct Answer
Option A
Solution
r‾com=m1r‾1+m2r‾2m1+m2{\overline r _{com}} = {{{m_1}{{\overline r }_1} + {m_2}{{\overline r }_2}} \over {{m_1} + {m_2}}}
=(1−9)i^+(2−6)j^+(1+3)k^4= {{(1 - 9)\widehat i + (2 - 6)\widehat j + (1 + 3)\widehat k} \over 4}
=−8i^−4j^+4k^4= {{ - 8\widehat i - 4\widehat j + 4\widehat k} \over 4}
r‾com=−2i^−j^+k^{\overline r _{com}} = - 2\widehat i - \widehat j + \widehat k
∣r‾∣=4+1+1=6\left| {\overline r } \right| = \sqrt {4 + 1 + 1} = \sqrt 6
∣i^+2j^+k^∣=6\left| {\widehat i + 2\widehat j + \widehat k} \right| = \sqrt 6
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