Problem 1:
Board A is placed on board B as shown. Both boards slide, without moving with respect to each other, along a frictionless horizontal surface at a speed v. Board B hits a resting board C "head-on." After the collision, boards B and C move together, and board A slides on top of board C and stops its motion relative to C in the position shown on the diagram. What is the length of each board? All three boards have the same mass, size, and shape. It is known there is no friction between boards A and B; the coefficient of kinetic friction between boards A and C is µk.
Solution:
 | Concepts, principles, relations
that apply to the problem: Momentum conservation, friction |
| Why do they apply? Board B and C collide inelastically. Board A slides over boards B and C, and we have sliding friction. |
| How do they apply? After the collision boards B and C stick together and move with speed v/2 in the lab frame (momentum conservation). There is no friction between A and B. Board A moves with speed v in the lab frame. The CM frame moves with velocity 2v/3 towards the right. In the CM frame board A moves with velocity v/3 towards the right and boards C and D move with velocity v/6 towards the left. Let us work in that frame. Let the right edge of the A at x' = 0 at t = 0, right after the collision. At time t, when the right edge of the board A is at x', the force on the board is the force of friction mN = mmg3x'/(2L) directed towards the left. L is the length of each board, 3x'/(2L) represents the fraction of the top board that is supported by the surface with friction. The acceleration is a = F/m, d2x'/dt2 = -mg3x'/(2L). The solution to this differential equation is x' = x'0sinwt, w = (3mg /(2L))1/2, v' = wx'0coswt, with the initial conditions x'(0) = 0, v'(0) = v/3 = wx'0. x'0 = v/(3w) = v'(0)/(3mg/(2L))1/2. When x' = x'0, v' = 0. x'0 therefore is the 2/3 of length of the board, x'0 = 2L/3. 2L/3 = v'(0)/(3mg/(2L))1/2, 2L2/9 = v'(0)2L/(3mg). The length of the board is L = 3v'(0)2/(2mg) = v2/(6mg) |
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Details of the calculation: |
Problem 2:
A rocket has an initial mass of m0 and a constant fuel burn rate of a. What is the minimum exhaust velocity that will allow the rocket to lift off from earth immediately after firing?
Solution:
| Concepts, principles, relations
that apply to the problem: Motion in one dimension, Newton's 2nd law, F = dp/dt, dp = p(t+dt) - p(t) |
| Why do they apply? Choose a coordinate axis with the positive direction upward. The system consists of the rocket body and the fuel. The speed of small amount of fuel dm changes by -v' in the time interval dt. In the same time interval the speed of the rocket and the remaining fuel increases by dv. The total momentum change of the system in the time interval dt therefore is dp = p(t+dt) - p(t) = (m-|dm|)(v+dv) + |dm|(v-v') - mv = mdv - |dm|v' (only first order terms are retained). |
| How do they apply? By Newton's 2nd law dp/dt is equal to the external force F = -mg. -mg = m(dv/dt) - (|dm/dt|)v'. Here |dm/dt| = -dm/dt = a. Therefore dv/dt = -g + av'/m. At t = 0 we have m = m0. We need dv/dt > 0 for the rocket to get off the ground, therefore we need v’ > m0g/a. |
| Details of the calculation: None |
Problem 3:
If two objects collide and one is initially at rest, is it possible for both to be at rest after the collision? Is it possible for one to be at rest after the collision? Explain!
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of momentum |
| Why do they apply? If a component of the total external force acting on a system is zero, then the corresponding component of the total momentum is conserved. |
| How do they apply? In collisions between two objects momentum is conserved. Since the initial momentum is not zero, the final momentum is not zero. Both objects cannot be at rest. It is possible for one of the objects to be at rest after the collision. For example, if the masses of the two objects are equal, then after a head-on elastic collision the object initially at rest is moving and the object initially moving is at rest. |
| Details of the calculation: None |
Problem 4:
A particle of mass m traveling with (non-relativistic) velocity u1 makes a head-on collision with a second particle of mass M, which is at rest in the laboratory. If the collision is completely inelastic, what fraction of the original kinetic energy remains after the collision?
Solution:
| Concepts, principles, relations
that apply to the problem: Inelastic collisions, momentum conservation |
| Why do they apply? In inelastic collisions: momentum is conserved, mechanical energy is not conserved. |
| How do they apply? Assume the particle is initially traveling in the positive x-direction. Before the collision: pi = mu1, Ei = ½mu12. After the collision: pf = (m+M)v = mu1. Ef = ½(M+m)v2 = ½[m2/(M+m)]u12. Ef/Ei = m/(M+m). |
| Details of the calculation: None |
Problem 5:
A rocket was launched vertically upwards. When it reached the highest point, it exploded into three equal mass pieces. One piece, which fell vertically down, was observed to take time t1 to reach the ground. The other two pieces both took time t2 to reach the ground. Find the height h(t1, t2 ) at which the rocket broke into three pieces.
Solution:
| Concepts, principles, relations
that apply to the problem: Freely falling objects, momentum conservation |
| Why do they apply? At the highest point the speed of the rocket is zero. The explosion gives the three pieces initial velocities. If the piece moving straight down has speed v, then the other two particle have upward component of velocity 0.5v and their horizontal components of velocity have equal magnitude and opposite sign.) |
| How do they apply? For the piece falling straight down: (i) 0 = h – vt1 – ½gt12. For the pieces moving initially upward with component of velocity ½v: (ii) 0 = h + ½vt2 – ½gt22. From (i): v = (h – ½gt12)/t1. Inserting into (ii): 0 = h + ½(h – ½gt12)t2/t1 – ½gt22. h + ½ht2/t1 = (1/4)gt1t2 + ½gt22 = ½gt2(½t1 + t2). h = ½gt2t1(½t1 + t2)/( t1 + ½t2). |
| Details of the calculation: None |
Problem 6:
A 15.2-g bullet hits a 0.463-kg block from below.
The initial speed of the bullet is 624 m/s and it emerges from the block at 131
m/s.
(a) How high does the block rise?
(b) If the block is 2.34-cm thick, estimate the average force on the block.
Assume that the bullet passes completely through before the block moves
appreciably.
Solution:
| Concepts, principles, relations
that apply to the problem: Momentum conservation, impulse |
| Why do they apply? In the collision between the bullet and the block momentum is conserved. The collision time is very short. |
| How do they apply? (a) m = 0.0152 kg , M = 0.463 kg, mvi = mvf + Mv v = m(vi – vf)/M = 0.0152*(489 m/s)/ 0.463 = 16 m/s The initial speed of the block is v = 16 m/s. It rises to a height h = v2/(2g) = 13.15 m. (b) The block receives an impulse Dp = Mv in the time interval Dt. Dt is the time it takes the bullet to pass through the block, Dt = 2*0.0234m/(vi + vf) = 6.23*10-5 s, assuming uniform deceleration. F = Dp/Dt = 1.19*105 N. |
| Details of the calculation: None |
Problem 7:
A uniform, dense rope of length b and mass per unit length m is coiled on a smooth table. You lift one end of the rope by hand vertically upward at constant speed u0. Find the force that you must apply to the rope when the end is a distance a above the table (a < b).
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's 2nd law, F = dp/dt |
| Why do they apply? Let a be the distance of the end of the rope above the table. Consider the rope to consist of sections of length dx. To move the rope upward with constant speed, you must apply an upward force equal to the force of gravity on the section of length a in addition to a force to change the momentum of a section dx from 0 to mdxu0 in a time interval dt. |
| How do they apply? F = mag + m(dx/dt)u0 = mag + mu02. |
| Details of the calculation: None |