Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Laws of Motion question

2020 · 5 Sep · Shift 2 · Q53
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Physics
  4. /Laws of Motion
  5. /2020 · 5 Sep · Shift 2 · Q53

Laws of Motion question

2020 · 5 Sep · Shift 2 · Q53

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A spaceship in space sweeps stationary interplanetary dust. As a result, its mass increases at a rate dM(t)dt{{dM\left( t \right)} \over {dt}}dtdM(t)​ = bv2(t), where v(t) is its instantaneous velocity. The instantaneous acceleration of the satellite is :
  1. A
    -bv3(t)
  2. B
    −2bv3M(t)- {{2b{v^3}} \over {M\left( t \right)}}−M(t)2bv3​
  3. C
    −bv3M(t)- {{b{v^3}} \over {M\left( t \right)}}−M(t)bv3​
  4. D
    −bv32M(t)- {{b{v^3}} \over {2M\left( t \right)}}−2M(t)bv3​
View written solutionFree

Correct answer: C

  1. Given data

A spaceship is moving in space and sweeps up stationary interplanetary dust. Its mass increases as

dMdt=bv2(t)\frac{dM}{dt} = b v^2(t)dtdM​=bv2(t)

We need the instantaneous acceleration a=dvdta = \dfrac{dv}{dt}a=dtdv​.

  1. Use conservation of momentum for variable mass system

In time interval dtdtdt:

  • spaceship mass changes from MMM to M+dMM+dMM+dM
  • velocity changes from vvv to v+dvv+dvv+dv
  • dust collected was initially at rest

Since there is no external force, total momentum is conserved.

Initial momentum:

Pi=MvP_i = MvPi​=Mv

Final momentum:

Pf=(M+dM)(v+dv)P_f = (M+dM)(v+dv)Pf​=(M+dM)(v+dv)

Because the dust was initially stationary, it contributed zero initial momentum.

So,

Mv=(M+dM)(v+dv)Mv = (M+dM)(v+dv)Mv=(M+dM)(v+dv)

Expanding and neglecting second-order small term dM dvdM\,dvdMdv:

Mv=Mv+M dv+v dMMv = Mv + M\,dv + v\,dMMv=Mv+Mdv+vdM

Thus,

M dv+v dM=0M\,dv + v\,dM = 0Mdv+vdM=0

Divide by dtdtdt:

Mdvdt+vdMdt=0M\frac{dv}{dt} + v\frac{dM}{dt} = 0Mdtdv​+vdtdM​=0

Hence,

dvdt=−vMdMdt\frac{dv}{dt} = -\frac{v}{M}\frac{dM}{dt}dtdv​=−Mv​dtdM​
  1. Substitute the given rate of mass increase

Given,

dMdt=bv2\frac{dM}{dt} = b v^2dtdM​=bv2

Therefore,

dvdt=−vM(bv2)\frac{dv}{dt} = -\frac{v}{M}(b v^2)dtdv​=−Mv​(bv2) dvdt=−bv3M\frac{dv}{dt} = -\frac{b v^3}{M}dtdv​=−Mbv3​

So the instantaneous acceleration is

−bv3(t)M(t)\boxed{-\frac{b v^3(t)}{M(t)}}−M(t)bv3(t)​​
  1. Option check
  • A: −bv3(t)-bv^3(t)−bv3(t) — missing division by mass, incorrect.
  • B: −2bv3M(t)-\dfrac{2bv^3}{M(t)}−M(t)2bv3​ — extra factor of 2, incorrect.
  • C: −bv3M(t)-\dfrac{bv^3}{M(t)}−M(t)bv3​ — correct.
  • D: −bv32M(t)-\dfrac{bv^3}{2M(t)}−2M(t)bv3​ — incorrect factor.
  1. Comparison with stored answer

Stored correct answer: C

Derived answer: C

They match.

PreviousNext

More from Laws of Motion

  • An insect is at the bottom of a hemispherical ditch of radius 1 m. It crawls up the ditch but starts slipping after it is at height h from the bottom. If the coefficient of friction between the ground and the insect is 0.75, then h is : (g…2020 · MCQ
  • A particle moving in the xy plane experiences a velocity dependent force F=k(vy​i+vx​j​), where vx and vy are the x and y components of its velocity v. If a…2020 · MCQ
  • A mass of 10 kg is suspended by a rope of length 4 m, from the ceiling. A force F is applied horizontally at the mid point of the rope such that the top half of the rope makes an angle of 45o with the vertical. Then F equal : (Take g = 10…2020 · MCQ
  • A block of mass 10 kg is kept on a rough inclined plane as shown in the figure. A force of 3 N is applied on the block. The coefficient of static friction between the plane and the block is 0.6. What should be the minimum value of force P,… Includes diagram2019 · MCQ
  • A mass of 10 kg is suspended vertically by a rope from the roof. When a horizontal force is applied on the rope at some point. the rope deviated at an angle of 45o at the roof point. If the suspended mass is at equilibrium, the magnitude…2019 · MCQ
  • A ball is thrown upward with an initial velocity V0 from the surface of the earth. The motion of the ball is affected by a drag force equal to m γ u2 (where m is mass of the ball, u is its instantaneous velocity and γ is a…2019 · MCQ
  • Two blocks A and B of masses mA = 1 kg and mB = 3 kg are kept on the table as shown in figure. The coefficient of friction between A and B is 0.2 and between B and the surface of the table is also 0.2. The maximum force F that can be… Includes diagram2019 · MCQ
  • A bullet of mass 20 g has an initial speed of 1 ms–1 , just before it starts penetrating a mud wall of thickness 20 cm. If the wall offers a mean resistance of 2.5 × 10–2 N, the speed of the bullet after emerging from the other side of the…2019 · MCQ