In the post dated 14-12-2008, two free-response question for practice was given to you.

As promised, I give below the answers along with the questions:

**(1) **ABCDEF is a track which is straight and horizontal in the region BCDE but curved in the regions AB and EF. The track is *smooth *everywhere except over a length 1.5 m in the horizontal region CD where the coefficient of kinetic friction is 0.2. A block of mass *m*_{1} = 2 kg is released from rest from position A which is at a height of 3.2 m. It slides down and collides with another stationary block of mass *m*_{2 }= 3 kg placed near position C (fig.). The velocity of the 3 kg mass *m*_{2} immediately after the collision is 4 ms^{–1}. Now answer the following:

(a) Determine the velocity of the 2 kg mass *m*_{1} immediately after the collision.

(b) State whether this collision is elastic or inelastic. Justify your answer.

(c) Calculate the kinetic energy lost by the 3 kg mass during its forward trip along the rough track CD.

(d) Determine the maximum height ‘*x*’* *up to which the 3 kg mass will rise along the smooth track EF.

(e) If the 2 kg mass *m*_{1} is removed (without disturbing any other thing) immediately after the collision, determine the distance from C where the 3 kg mass will finally come to rest.

**(a)** The speed *u*_{1}* *of the mass *m*_{1} just before it collides with the mass *m*_{2} is given by

*m*_{1}*gh* = ½ *m*_{1 }*u*_{1}^{2}, on equating the loss of gravitational ^{ }potential energy to the gain of kinetic energy.

Therefore, _{ }*u*_{1} = √(2*gh*) _{ }= √(2×10×3.2) = 8 ms^{–1}.

The initial momentum of the system (of masses *m*_{1} and* m*_{2}) is (*m*_{1}*u*_{1} + 0) and the final momentum is (*m*_{1}*v*_{1} +* m*_{2}*v*_{2}) where *v*_{1} and *v*_{2} are the velocities of masses *m*_{1} and* m*_{2} respectively after the collision.

Since the momentum is conserved, *m*_{1}*u*_{1} = *m*_{1}*v*_{1} +* m*_{2}*v*_{2}.

Substituting known values, 2×8 = 2* v*_{1} + 3×4 from which *v*_{1} = 2 ms^{–1}.

**(b)** This collision is inelastic since the *kinetic energy* is not conserved as shown below:

Initial kinetic energy = ½ *m*_{1}*u*_{1}^{2} + 0 = ½ ×2×8^{2} = 64 J

Final kinetic energy = ½ *m*_{1}*v*_{1}^{2} + ½ *m*_{2}*v*_{2}^{2} = ½ (2×2^{2} + 3×4^{2}) = (4+24) J = 28 J, which is less than the initial kinetic energy.

**(c)** The kinetic energy lost by the 3 kg mass during its forward trip along the rough track CD is *μ*_{k}*m*_{2}*gd = *0.2×3×10×1.5 = 9 J (since the coefficient of kinetic friction (*μ*_{k}) is 0.2 and the distance moved (*d*)* *is 1.5 m.

**(d)** The kinetic energy remaining in the 3 kg mass after its forward trip along CD is (24 J – 9 J) = 15 J. The maximum height ‘*x*’* *up to which the 3 kg mass will rise along the smooth track EF is given by

*m*_{2}*gx = *15 J* *

Therefore, *x* = 15/(3×10) = 0.5 m.

**(e**) Since the mass *m*_{1} is removed after the first collision there is no possibility for further collisions. The mass *m*_{2} comes down and moves backwards from D to E losing another 9 J of energy. The kinetic energy remaining in it is now 6 J only and rises up along the smooth track CBA and returns after reaching the maximum possible height. It will stop at a point distant ‘*s*’ from C after losing its energy in doing work against traction. *s* is evidently given by

*μ*_{k}*m*_{2}*gs =*6 J

Therefore *s = *6/(0.2×3×10) = 1 m.

**(2)**An object A of mass 5 kg moving along the positive x-direction has displacement

*x*given by

*x =*0.2

*t*+ 0.18

*t*

^{2}. The force acting on the object ceases after 5 seconds and the object moves with uniform velocity. Another object B of mass 10 kg moving in the positive y-direction has kinetic energy equal to the potential energy of a spring of force constant 3000 Nm

^{–1}compressed through 0.1 m from its natural length. The objects A and B moving with their respective constant velocities collide at the origin O and stick together.

Now answer the following questions:

(a) Determine the momentum and kinetic energy of object A just before collision.

(b) Determine the momentum and kinetic energy of object B just before collision

(c) Calculate the velocity of the combined mass (after A and B stick together).

(d) Calculate the change in kinetic energy of the system because of the collision.

(e) Is this collision elastic or inelastic? Justify your answer.

**(a) **The velocity v_{A} of object A will be constant after 5 s since the force ceases to act. This is obtained by substituting *t = *5 s in* *dx/dt:

Therefore, v_{A} = 0.2 + 2×0.18*t = *0.2 + 2×0.18×5 = 2 ms^{–1} * *

Momentum of object A is P_{A} =_{ } *m*_{A}v_{A} = 5×2 = **10 kg**** ms**^{–1}* *

Kinetic energy of object A is *K*_{A} = P_{A}^{2}/2*m*_{A} = 100/10 = **10 J**

[Or, *K*_{A} = ½ *m*_{A} v_{A}^{2}, if you would prefer this form].

**(b)** *kx*^{2} where *k* is the spring constant and *x* is the compression. Substituting appropriate values, the energy is ½ ×3000×(0.1)^{2} = 15 J. Since the kinetic energy *K*_{B} of the object B is equal to the above energy, we have

*K*_{B} = **15 J**.

Therefore, P_{B}^{2}/2*m*_{B} = 15 from which the momentum P_{B} of object B is obtained as

P_{B} = √(2×10×15) = **10****√3 kg ms**^{–1}

**(c) **The momentum vectors P_{A} and P_{B} are shown in the figure. The resultant momentum P gives the momentum of the combined mass and is given by

P = √(P_{A}^{2} + P_{B}^{2}) = √(100+300) = 20 kg ms^{–1}

The velocity ‘v’ of the combined mass is given by

v = P/(*m*_{A}*+m*_{B}) = 20/15 = **1.333 ****ms**^{–1}

The velocity is directed as shown, making an angle *θ* with the x-axis, given by

tan*θ* = P_{B}/P_{A} = (10√3)/10 = √3 so that *θ* = 60**º**

**(d)** Kinetic energy *K *of the combined mass after the collision is given by

*K* = P^{2}/2(*m*_{A}*+m*_{B}) = 20^{2}/2(5+10) = 13.333 J

The initlal kinetic energy of the system is *K*_{A}* *+ *K*_{B} = 10+15 = 25 J.

Thefefore, the change in kinetic energy of the system because of the collision is 13.333 –25 = **–11.667 J**

[Negative sign shows that the kinetic energy has decreased].

**(e) **The collision is *inelastic* since the kinetic energy is not conserved.

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