A body of mass 2m moving with velocity v makes a head on elastic

Subject

Physics

Class

JEE Class 12

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 Multiple Choice QuestionsMultiple Choice Questions

1.

A person observes that the full length of a train subtends an angle of 15°. If the distance between the train and the person is 3 km, the length of the train, calculated using parallax method, in meters is

  • 45

  • 45 π

  • 250 π

  • 75 π


2.

In a measurement, random error

  • can be decreased by increasing the number of readings and averaging them

  • can be decreased by changing the person who takes the reading

  • can be decreased by using new instrument

  • can be decreased by using a different method in taking the reading


3.

In order to measure the period of a single pendulum using a stop clock, a student repeated the experiment for 10 times and noted down the time period for each experiment as 5.1, 5.0, 4.9, 4.9, 5.1, 5.0, 4.9, 5.1, 5.0, 4.9 s. The correct way of expressing the result for the period is

  • 4.99 s

  • 5.0 s

  • 5.00 s

  • 4.9 s


4.

The following figure gives the movement of an object. Select the correct statement from the given choices.

       

  • The total distance travelled by the object is 975 m

  • The maximum acceleration of the object is 2m/s2

  • The maximum deceleration happend between 25th and 85th seconds

  • The object was at rest between 10th and 15th seconds


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5.

Two object P and Q, travelling in the same direction starts from rest. While the object P starts at time t = 0 and object Q starts later at t = 30 min. The object P has an acceleration of 40 km/h2. To catch P at a distance of 20 km, the acceleration of Q should be

  • 40 km/h2

  • 80 km/h2

  • 160 km/h2

  • 120 km/h2


6.

A train of length L move with a constant speed Vt. A person at the back of the train fires a bullet at time t = 0 towards a target which is at a distance of D (at time t = 0) from the front of the train (on the same direction of motion). Another person at the front of the train fires another bullet at time t = T towards the same target. Both bullets reach the target at the same time. Assuming the speed of the bullets Vb are same, the length of the train is

  • T × (Vb × 2Vt)

  • T × (Vb + Vt)

  • 2VbDVb + Vt + T Vb - Vt

  • T × (Vb − 2Vt)


7.

From the ground, a projectile is fired at an angle of 60 degrees to the horizontal with a speed of 20 m/s. Take, acceleration due to gravity as 10 m/s2 . The horizontal range of the projectile is

  • 103 m

  • 20 m

  • 203

  • 403 m


8.

A person from a truck, moving with a constant speed of 60 km/h, throws a ball upwards with a speed of 60 km/h. Neglecting the effect of Earth and choose the correct answer from the given choice.

  • The person cannot catch the ball when it comes down since the truck is moving

  • The person can catch the ball when it comes down, if the truck is stopped immediately after throwing the ball

  • The person can catch the ball when it comes down, if the truck continues to move with a constant speed of 60 km/h

  • The person can catch the ball when it comes down, if the truck moves with speed more than 60 km/h


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9.

A body of mass 2m moving with velocity v makes a head on elastic collision with another body of mass m which is initially at rest. Loss of kinetic energy of the colliding body (mass 2m) is

  • 1/9  of its initial kinetic energy

  • 1/6  of its initial kinetic energy 

  • 8/9 of its initial kinetic energy

  • 1/2 of its initial kinetic energy


C.

8/9 of its initial kinetic energy

Initial K.E of ball of mass 2m = K1

        = 12 × 2m × v2= mv2

Collision is elastic so both K.E and momentum are conserved. Let velocities of balls are v1 and v2 after collision

       

So, KE is conserved

 12 2mv2 = 12 2mv12 + 12 mv22  v2 = v12 + 12 v22              ......... (i)

And, momentum is conserved

⇒ (2m)v + m(0) = 2m (v1) + mv2

⇒                  2v = 2v1 + v2        ........ (ii)

Put this value in Eq. (i), we get

         v2 = v12 + 12 × 4 v - v12 3v12 - 4vv1 + v2 = 0 3 v1v2 - 4 v1v + 1 = 0or    v1v = - - (-4) ± 16 - 122 × 3 v1v  = 4 ± 22 × 3    v1 = v (Not possible)or      v1 = 13 v

So, final K.E of ball of mass 2m,

   k2 = 12 2m v12 = 12 × 2m × v29 = 19 k1

Hence, loss of K.E. of 1st ball

    = K1 - 89 K1 = 89 K1


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10.

Displacement x (in meters), of body of mass 1 kg as a function of time t, on a horizontal smooth surface is given as x = 2t2 . The work done in the first one second by the external force is

  • 1 J

  • 2 J

  • 4 J

  • 8 J


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