Practice Start Test

Center of Mass and Collision

NEET Physics · 98 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
A person holding a rifle (mass of person and rifle together is 100 kg) stands on a smooth surface and fires 10 shots horizontally, in 5 s. Each bullet has a mass of 10 g with a muzzle velocity of 800 m s −-1 . The final velocity acquired by the person and the average force exerted on the person are :
A −-0.08 ms −-1 , 16 N
B −-0.8 ms −-1 , 8 N
C −-1.6 ms −-1 , 16 N
D −-1.6 ms −-1 , 8 N
Correct Answer
Option B
Solution

According to law of conservation of momentum MV + mnv = 0

⇒V−−mNvM−−0.01 kg×10×800 m/s100\Rightarrow V - {{ - mNv} \over M} - {{ - 0.01\,kg \times 10 \times 800\,m/s} \over {100}}
⇒−0.8 m/s\Rightarrow - 0.8\,m/s

According to work energy theorem, Average work done = Change in average kinetic energy i.e,

Fav×Sav=12mVrms2{F_{av}} \times {S_{av}} = {1 \over 2}mV_{rms}^2
⇒FavVmax⁡t2=12mVrms22\Rightarrow {{{F_{av}}{V_{\max }}t} \over 2} = {1 \over 2}m{{V_{rms}^2} \over 2}
⇒Fav=8N\Rightarrow {F_{av}} = 8N
Q22
An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass 1 kg moves with a speed of 12 m s −-1 and the second part of mass 2 kg moves with 8 m s −-1 speed. If the third part files off with 4 m s −-1 speed, then its mass is :
A 7 kg
B 17 kg
C 3 kg
D 5 kg
Correct Answer
Option D
Solution

The situation is as shown in the figure. According to law of conservation of linear momentum

p→1+p→2+p→3=0{\overrightarrow p _1} + {\overrightarrow p _2} + {\overrightarrow p _3} = 0

∴\therefore

p→3=−(p→1+p→2){\overrightarrow p _3} = - \left( {{{\overrightarrow p }_1} + {{\overrightarrow p }_2}} \right)

Here,

p→1=(1 kg)(12 ms−1)i^=12i^ kg ms−1{\overrightarrow p _1} = (1\,kg)(12\,m{s^{ - 1}})\widehat i = 12\widehat i\,kg\,m{s^{ - 1}}
p→2=(2 kg)(8 ms−1)j^=16i^ kg ms−1{\overrightarrow p _2} = (2\,kg)(8\,m{s^{ - 1}})\widehat j = 16\widehat i\,kg\,m{s^{ - 1}}
∴p→3=−(12i^+16j^) kgms−1\therefore {\overrightarrow p _3} = - \left( {12\widehat i + 16\widehat j} \right)\,kgm{s^{ - 1}}

The magnitude of p 3 is

p3=(12)2+(16)2=20 kgms−1{p_3} = \sqrt {{{\left( {12} \right)}^2} + {{\left( {16} \right)}^2}} = 20\,kgm{s^{ - 1}}
∴m3=p3v3=20 kgms−14ms−1=5kg\therefore {m_3} = {{{p_3}} \over {{v_3}}} = {{20\,kgm{s^{ - 1}}} \over {4m{s^{ - 1}}}} = 5kg
Q23
Three masses are placed on the x-axis : 300 g at origin, 500 g at x = 40 cm and 400 g at x = 70cm. The distance of the centre of mass from the orgin is :
A 40 cm
B 45 cm
C 50 cm
D 30 cm
Correct Answer
Option A
Solution

The distance of the centre of mass of the system of three masses from the origin O is

XCM=m1x1+m2x2+m3x3m1+m2+m3{X_{CM}} = {{{m_1}{x_1} + {m_2}{x_2} + {m_3}{x_3}} \over {{m_1} + {m_2} + {m_3}}}
=300×0+500×40+400×70300+500+400= {{300 \times 0 + 500 \times 40 + 400 \times 70} \over {300 + 500 + 400}}
=500×40+400×701200=400[50+70]1200= {{500 \times 40 + 400 \times 70} \over {1200}} = {{400\left[ {50 + 70} \right]} \over {1200}}
=50+703=1203=40 cm= {{50 + 70} \over 3} = {{120} \over 3} = 40\,cm
Q24
Two spheres A and B of masses m 1 and m 2 respectively collide. A is at rest initially and B is moving with velocity v along x-axis. After collision B has a velocity v2{v \over 2} in a direction perpendicular to the original direction. The mass A moves after collision in the direction :
A same as that of B
B opposite to that of B
C θ\theta = tan −-1 (12)\left( {{1 \over 2}} \right) to the x-axis
D θ\theta = tan −-1 (−12)\left( { - {1 \over 2}} \right) to the x-axis
Correct Answer
Option D
Solution

conservation of linear momentum along x-direction

m2v=m1vx⇒m2vm1=vx{m_2}v = {m_1}{v_x} \Rightarrow {{{m_2}v} \over {{m_1}}} = {v_x}

along y-direction

m2×v2=m1vy⇒vy=m2v2m1{m_2} \times {v \over 2} = {m_1}{v_y} \Rightarrow {v_y} = {{{m_2}v} \over {2{m_1}}}

Note: Let A moves in the direction, which makes an angle q with initial direction i.e.

tan⁡θ=vyvx=m2v2m1/m2v2m1\tan \theta = {{{v_y}} \over {{v_x}}} = {{{m_2}v} \over {2{m_1}}}/{{{m_2}v} \over {2{m_1}}}
tan⁡θ=12\tan \theta = {1 \over 2}
tan⁡θ=12⇒θ=tan⁡−1(12)\tan \theta = {1 \over 2} \Rightarrow \theta = {\tan ^{ - 1}}\left( {{1 \over 2}} \right)

to the x-axis.

Q25
Two persons of masses 55 kg and 65 kg respectively, are at the opposite ends of a boat. The length of the boat is 3.0 m and weights 100 kg. The 55 kg man walks up to the 65 kg man and sits with him. If the boat is in still water the centre of mass of the system shifts by :
A 3.0 m
B 2.3 m
C zero
D 0.75 m
Correct Answer
Option C
Solution

As no external force acts on the system, therefore centre of mass will not shift.

Q26
Two particles which are initially at rest, move towards each other under the action of their internal attraction. If their speeds are v and 2v at any instant, then the speed of centre of mass of the system will be :
A 2v
B zero
C 1.5 v
D v
Correct Answer
Option B
Solution

If no external force actson a system of particles, the centre of mass remains at rest.

So, speed of centre of mass is zero.

Q27
An explosion blows a rock into three parts. Two parts go off at right angles to each other. These two are, 1 kg first part moving with a velocity of 12 m s −-1 and 2 kg second part moving with a velocity 8 m s −-1 . If the third part flies off with a velocity of 4 m s −-1 , its mass would be :
A 7 kg
B 17 kg
C 3 kg
D 5 kg
Correct Answer
Option D
Solution

When an explosion breaks a rock, by the law of conservation of momentum, initial momentum is zero and for the three pieces, Total momentum of the two pieces 1 kg and 2 kg

=122+162=20 kgms−1= \sqrt {{{12}^2} + {{16}^2}} = 20\,kgm{s^{ - 1}}

The third piece has the same momentum and in the direction opposite to the resultant of these two momentum.

∴\therefore Momentum of the third piece = 20 kg ms –1 Velocity = 4 ms –1 ∴\therefore Mass of the 3 rd piece =

mvv=204{{mv} \over v} = {{20} \over 4}

= 5 kg

Q28
Two bodies of mass 1 kg and 3 kg have position vectors i^+2j^+k^\widehat i + 2\widehat j + \widehat k and −3i^−2j^+k^ - 3\widehat i - 2\widehat j + \widehat k, respectively. The center of mass of this system has a position vector :
A −2i^−j^+k^ - 2\widehat i - \widehat j + \widehat k
B 2i^−j^−2k^2\widehat i - \widehat j - 2\widehat k
C −i^+j^+k^ - \widehat i + \widehat j + \widehat k
D −2i^+2k^ - 2\widehat i + 2\widehat k
Correct Answer
Option A
Solution

The position vector of the centre of mass of two particle system is given by

R→=m1R1→+m2R2→(m1+m2)\overrightarrow R = {{{m_1}\overrightarrow {{R_1}} + {m_2}\overrightarrow {{R_2}} } \over {\left( {{m_1} + {m_2}} \right)}}
=14[−8i^−4j^+4k^]=−2i^−j^+k^= {1 \over 4}\left[ { - 8\widehat i - 4\widehat j + 4\widehat k} \right] = - 2\widehat i - \widehat j + \widehat k
Q29
A shell of mass 200 gm is ejected from a gun of mass 4 kg by an explosion that generates 1.05 kJ of energy. The initial velocity of the shell is :
A 40 ms −-1
B 120 ms −-1
C 100 ms −-1
D 80 ms −-1
Correct Answer
Option C
Solution

As per laws of conservation of momentum and energy, if v and V represents initial velocity of shell of mass m and recoil velocity of gun of mass M respectively, then mv + MV = 0

V=−(mM)vV = - \left( {{m \over M}} \right)v
=−(0.24)v=−v20m/s= - \left( {{{0.2} \over 4}} \right)v = - {v \over {20}}m/s

Here (–) sign shows that recoil velocity of gun is opposite to velocity of shell.

The energy of explosion is used to impart kinetic energy to shell and gun so that we have

12{1 \over 2}

(mv 2 + MV 2 ) = 1.05 × 10 3

12{1 \over 2}

[0.2v 2 + (4v 2 / 400)] = 1.05 ×10 3 ⇒\Rightarrow v 2 = 10 4 ⇒\Rightarrow v = 100 m/s

Q30
A particle of mass m is projected with velocity v making an angle of 45 o with the horizontal. When the particle lands on the level ground the magnitude of the change in its momentum will be :
A mv2mv\sqrt 2
B zero
C 2mv
D mv / 2\sqrt 2
Correct Answer
Option A
Solution

The magnitude of the resultant velocity at the point of projection and the landing point is same.

Clearly, change in momentum along horizontal (i.e along x-axis) = mvcosθ\theta – mvcosθ\theta = 0 Change in momentum along vertical (i.e. along y–axis) = mv sinθ\theta – (–mv sinθ\theta) = 2 mvsinθ\theta = 2mv × sin 45° =

2mv×12=2mv2mv \times {1 \over {\sqrt 2 }} = \sqrt 2 mv

Hence, resultant change in momentum =

2mv\sqrt 2 mv
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