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`“Life is like riding a bicycle.  To keep your balance you must keep moving.”–Albert Einstein`

## Wednesday, December 17, 2008

### Answers to Fre-response Pracice Questions on AP Physics B&C Work, Energy & Power

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 m1 = 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 m2 = 3 kg placed near position C (fig.). The velocity of the 3 kg mass m2 immediately after the collision is 4 ms–1. Now answer the following:

(a) Determine the velocity of the 2 kg mass m1 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 m1 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 u1 of the mass m1 just before it collides with the mass m2 is given by

m1gh = ½ m1 u12, on equating the loss of gravitational potential energy to the gain of kinetic energy.

Therefore, u1 = √(2gh) = √(2×10×3.2) = 8 ms–1.

The initial momentum of the system (of masses m1 and m2) is (m1u1 + 0) and the final momentum is (m1v1 + m2v2) where v1 and v2 are the velocities of masses m1 and m2 respectively after the collision.

Since the momentum is conserved, m1u1 = m1v1 + m2v2.

Substituting known values, 2×8 = 2 v1 + 3×4 from which v1 = 2 ms–1.

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

Initial kinetic energy = ½ m1u12 + 0 = ½ ×2×82 = 64 J

Final kinetic energy = ½ m1v12 + ½ m2v22 = ½ (2×22 + 3×42) = (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 μkm2gd = 0.2×3×10×1.5 = 9 J (since the coefficient of kinetic friction (μk) 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

m2gx = 15 J

Therefore, x = 15/(3×10) = 0.5 m.

(e) Since the mass m1 is removed after the first collision there is no possibility for further collisions. The mass m2 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

μkm2gs =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 t2. 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. (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.

(a) The velocity vA 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, vA = 0.2 + 2×0.18t = 0.2 + 2×0.18×5 = 2 ms–1

Momentum of object A is PA = mAvA = 5×2 = 10 kg ms–1

Kinetic energy of object A is KA = PA2/2mA = 100/10 = 10 J

[Or, KA = ½ mA vA2, if you would prefer this form].

(b) Potential energy of a compressed spring is ½ kx2 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 KB of the object B is equal to the above energy, we have

KB = 15 J.

Therefore, PB2/2mB = 15 from which the momentum PB of object B is obtained as

PB = √(2×10×15) = 10√3 kg ms–1 (c) The momentum vectors PA and PB are shown in the figure. The resultant momentum P gives the momentum of the combined mass and is given by

P = √(PA2 + PB2) = √(100+300) = 20 kg ms–1

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

v = P/(mA+mB) = 20/15 = 1.333 ms–1

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

tanθ = PB/PA = (10√3)/10 = √3 so that θ = 60º

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

K = P2/2(mA+mB) = 202/2(5+10) = 13.333 J

The initlal kinetic energy of the system is KA + KB = 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.