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A trolley of mass 200 kg moves with a uniform speed of 36 km/h on a frictionless track. A child of mass 20 kg runs on the trolley from one end to the other (10 m away) with a speed of 4 m s–1 relative to the trolley in a direction opposite to the its motion, and jumps out of the trolley. What is the final speed of the trolley? How much has the trolley moved from the time the child begins to run?

Given,

Mass of the trolley, 
M = 200 kg 

Speed of the trolley, v = 36 km/h = 10 m/s 

Mass of the boy, m = 20 kg 

Initial momentum of the system of the boy and the trolley, 

= (M + m)v 

= (200 + 20) × 10 

= 2200 kg m/s 

Let v' be the final velocity of the trolley with respect to the ground. 

Final velocity of the boy with respect to the ground = v' - 4 

Final momentum = Mv' + m(v' - 4) 
 
                           = 200v' + 20v' - 80 

                           = 220v' – 80 

As per the law of conservation of momentum,
 

Initial momentum = Final momentum 

                   2200 = 220v' – 80 

∴                       v' =

Length of the trolley, l = 10 m 

Speed of the boy, v'' = 4 m/s 

Time taken by the boy to run, t = = 2.5 s
∴ Distance moved by the trolley = v'' × 

                                                = 10.36 × 2.5

                                                = 25.9 m
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A bolt of mass 0.3 kg falls from the ceiling of an elevator moving down with an uniform speed of 7 m s–1. It hits the floor of the elevator (length of the elevator = 3 m) and does not rebound. What is the heat produced by the impact? Would your answer be different if the elevator were stationary?

Given,

Mass of the bolt, 
m = 0.3 kg 

Speed of the elevator = 7 m/s 

Height, h = 3 m 

Since the relative velocity of the bolt with respect to the lift is zero, at the time of impact, potential energy gets converted into heat energy. 

Heat produced = Loss of potential energy  

                      = mg

                      = 0.3 × 9.8 × 3 

                      = 8.82 J 

The heat produced will remain the same even if the lift is stationary. This is because of the fact that the relative velocity of the bolt with respect to the lift will remain zero.
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A 1 kg block situated on a rough incline is connected to a spring of spring constant 100 N m–1 as shown in Fig. 6.17. The block is released from rest with the spring in the unstretched position. The block moves 10 cm down the incline before coming to rest. Find the coefficient of friction between the block and the incline. Assume that the spring has a negligible mass and the pulley is frictionless.


Given,

Mass of the block, 
m = 1 kg 

Spring constant, k = 100 N m–1 

Displacement in the block, x = 10 cm = 0.1 m

The given situation can be illustrated in the following figure, 



At equilibrium, 

Normal reaction, R = mg cos 37° 

Frictional force, f = μ R

                          = mg Sin 37

where, μ is the coefficient of friction

Net force acting on the block = mg sin 37° – 

                                            = mgsin 37° – μmgcos 37° 

                                            = mg(sin 37° – μcos 37°) 

At equilibrium, the work done by the block is equal to the potential energy of the spring.

i.e., 
mg(sin 37° – μcos 37°)x =  kx2
 1 × 9.8 (Sin 370 - μcos 37°) = × 100 × (0.1) 

                0.602 - μ × 0.799 = 0.510 

∴                        μ =   =  0.115 
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Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track (Fig. 6.16). Will the stones reach the bottom at the same time? Will they reach there with the same speed? Explain. Given θ1 = 30°, θ2 = 60°, and = 10 m, what are the speeds and times taken by the two stones?


The given question can be illustrated using the figure below: 

 


AB and AC are two smooth planes inclined to the horizontal at ∠θ1 and ∠θ2 respectively.

The height of both the planes is the same, therefore, both the stones will reach the bottom with same speed.

As P.E. at O = K.E. at A = K.E. at B 

Therefore, 

     mgh = 1/2 mv12 = 1/2 mv2

∴                          v1 = v

As it is clear from fig. above, acceleration of the two blocks are

 
a1 = g sin θ1 ,

 a2 = g sin θ

As θ2 > θ

∴       a2 > a

From v = u + at 

            = 0 + 
at 

      t = v/

As ∝ 1/a, and a2 > a

∴    t2 < t

That is, the second stone will take lesser time and reach the bottom earlier than the first stone.

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Which of the following potential energy curves cannot possibly describe the elastic collision of two billiard balls? Here r is the distance between centers of the balls, R is the radius of each ball. 


If collisions are elastic collisions then potential energy is function of position only after the two colliding bodies just make the contact with each other and potential energy is zero when the two bodies are separate. The potential energy increases with decrease in separation after two bodies make the contact with each other,
i.e.            straight V equals 0                                                           for straight r greater or equal than 2 straight R
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∴              Possible graph is,
                  
If collisions are elastic collisions then potential energy is functio
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