Problem 1:
After a completely inelastic collision between two objects of equal mass, each having initial speed v, the two move off together with speed v/3. What was the angle between their initial directions?
Solution:
 | Concepts, principles, relations that apply to the
problem: Conservation of momentum |
| Why do they apply? The external force acting on the system is zero, the total momentum is conserved. |
| How do they apply? p1x + p2x = pfx = pf, p1y + p2y = 0. 2mv cosθ = 2mv/3, cosθ = 1/3, θ = 70.5o. The angle between their initial directions is 2θ = 141o. |
| Details of the calculation: None |
Problem 2:
A block of mass m1 and initial velocity v1
collides head-on with a stationary block of mass m2. The mass m2
compresses a spring of spring constant k. Neglect friction and assume the
collision is elastic.
(a) What is the velocity v2 of m2 just after the
collision?
(b) What is the maximum compression of the spring?
Solution:
| Concepts, principles, relations
that apply to the problem: Elastic collisions |
| Why do they apply? We are asked to analyze an elastic collision between two masses. |
| How do they apply? (a) In elastic collisions energy and momentum are conserved. Let v1and v2 be the velocities of m1 and m2 just after the collision. Momentum conservation: m1v1 = m1 v1 + m2 v2. Energy conservation: m1v12 = m1 v12 + m2 v22. v2 = 2v1/(1 + m2/m1). (b) ½m2v22 = ½kd2, where d is the maximum compression of the spring d = (m2/k)1/22v1/(1 + m2/m1). |
| Details of the calculation: None |
Problem 3:
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 Δp = Mv in the time interval Δt. Δt is the time it takes the bullet to pass through the block, Δt = 2*0.0234m/(vi + vf) = 6.23*10-5 s, assuming uniform deceleration. F = Δp/Δt = 1.19*105 N. |
| Details of the calculation: None |
Problem 4:
A neutron in a reactor makes an elastic head-on collision with the nucleus of
a carbon atom initially at rest.
(a) What fraction of the neutron's kinetic energy is transferred to the carbon
nucleus?
(b) If the initial kinetic energy of the neutron is 1.6*10-13
J, find its final kinetic energy and the kinetic energy of the carbon
nucleus after the collision.
(The mass of the carbon nucleus is about 12 times the mass of the neutron.)
Solution:
| Concepts, principles, relations that apply to the
problem: Elastic collisions, momentum conservation, energy conservation |
| Why do they apply? As stated in the problem, the collision of the two particles is elastic. |
| How do they apply? Momentum conservation: (i) m1v1i = m1v1f + m2v2f. Energy conservation: (ii) (1/2)m1v1i2 = (1/2)m1v1f2 + (1/2)m2v2f2. |
| Details of the calculation: (a) We can solve this system of two equations for the ratio v2f/v1i. We obtain v2f2 = (m1/m2)(v1i2 - v1f2) from (ii), v1f = v1i - (m2/m1)v2f from (i). We can therefore write v1f2 = v1i2 + (m2/m1)2v2f2 - 2(m2/m1)v1iv2f. v2f2 = (m1/m2)( 2(m2/m1)v1iv2f - (m2/m1)2v2f2) = 2v1iv2f - (m2/m1)v2f2. (1 + (m2/m1))v2f = 2v1i. v2f = 2m1v1i/(m1+m2). v2f/v1i = 2m1/(m1+m2) = 2/13 = 0.15. The fraction of the kinetic energy transferred to the carbon nucleus is (m2/m1)(v2f/v1i)2 = 12(0.15)2 = 0.284. (b) The initial kinetic energy of the of the neutron is 160 fJ and the final kinetic energy of the carbon nucleus is 45.4 fJ. (1 femtoJoule = 1 fJ = 10-15 J.) The final kinetic energy of the neutron is (160 - 45.4) fJ = 114.6 fJ. |
Problem 5:
A rope of length l slides over the edge of a table. Initially a piece x0
of it hangs without motion over the side of the table. Let x be the length of
rope hanging vertically at time t. The rope is assumed to be perfectly
flexible.
Show that E = T + U is a constant of motion.
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of mechanical energy | ||
| Why do they apply? We are asked to show that the mechanical energy is conserved in a given situation.
| ||
| How do they apply? Let l be the length of the rope, ρ its mass per unit length, and let x be the part hanging over the edge.  This is an effectively one-dimensional problem. The picture below shows a blowup of the corner, the section of the rope bending around the corner has infinitesimal length.  Fg - T = d(ρxv)/dt for the vertical part of the rope, T = d(ρ(l-x)v)/dt for the horizontal part of the rope. Defining p = ρlv = ρldx/dt we have Fg = dp/dt = ρld2x/dt2 = ρgx. d2x/dt2 = gx/l. T = (1/2)ρl(dx/dt)2, U = -ρg(x2/2), E = T + U = (1/2)ρl(dx/dt)2 - ρg(x2/2) dE/dt = ρl(dx/dt)d2x/dt2 - ρgxdx/dt = ρgxdx/dt - ρgxdx/dt = 0. | ||
| Details of the calculation: None |
Problem 6:
A chain lies pushed together at the edge of a table, except for a piece
which hangs over it, initially at rest. The links of the chain start
moving, one at a time.
Show that E = T + U is not a constant of
motion.
Solution:
| Concepts, principles, relations
that apply to the problem: Conservation of mechanical energy | ||
| Why do they apply? We are asked to show that the mechanical energy is not conserved in a given situation.
| ||
| How do they apply? Let ρ its mass per unit length of the chain. Let l be the length of the chain that hangs over the edge. This part of the chain is moving. Let the x-axis point downward and let the tabletop be at x = 0. px = ρlv = ρldl/dt. Fx = dpx/dt = ρ(dl/dt)2 + ρld2l/dt2 = ρlg. T = (1/2)ρlv2 = (1/2)ρl(dl/dt)2, U = -ρlg(l/2), E = T + U = (ρl/2)[(dl/dt)2 - gl]. dE/dt = (ρ/2)(dl/dt)3 + ρl(dl/dt)d2l/dt2 - ρgldl/dt. d2l/dt2 = g - (1/l)(dl/dt)2, from Fx = dpx/dt. dE/dt = (ρ/2)(dl/dt)3 + ρlg(dl/dt) - ρ(dl/dt)3- ρgldl/dt = -(ρ/2)(dl/dt)3 = -(ρ/2)v3 ≠ 0. |
Problem 7:
A uniform heavy chain of mass
λ
per unit length hangs vertically so that the low end just touches a
horizontal table. The upper end is released and the chain falls on the table.
Find the force the chain exerts on the table after it has fallen a distance x.
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's 2nd law, F = dp/dt |
| Why do they apply? Consider the chain to consist of sections of length dx. After the chain has fallen a distance x, the sections that have not yet reached the table have velocity v = (2gx)1/2. The table has to support a section of the chain of length x and has to reduce the momentum of a section of length dx from λdxv to zero in a time interval dt. |
| How do they apply? The table has to exert a force F = λxg + λv2 in the upward direction. F = λxg + λ2gx = 3λxg. |
| Details of the calculation: None |
Problem 8:
A chain of mass m and length L rests with (1 - α)L of its length on a table top and αL of its length hanging over the smooth edge, as shown in the following figure.
(a) The coefficient of friction
of the tabletop is μ.
What is the maximum value, αc, for which the
chain remains stationary?
(b) If α
is larger than αc, when released the chain will slide off
the table. What is the velocity of the chain when the last link leaves the
table?
Solution:
| Concepts, principles, relations
that apply to the problem: Newton's 2nd law: F = dp/dt, Work-kinetic energy theorem: ∫F·dr = ΔT. |
| Why do they apply? This is an effectively one-dimensional problem. The forces acting on the chain are gravity, the normal force, and friction. ∫F·dr = (1/2)mv2. |
| How do they apply? (a) ρ = m/L = linear mass density, ραcLg - ρ(1-αc)Lgμ = 0, gravity frictional force αc - (1-αc)μ = 0, αc = μ/(μ+1). |
| Details of the calculation: (b)  |
Problem 9:
A spherical water droplet falls without friction under the influence of gravity through an atmosphere saturated with water vapor. Let its initial radius at t = 0 be C, its initial velocity v0. As a result of condensation, the water drop experiences a continuous increase in mass, proportional to its surface. Its radius R then increases linearly with time. Integrate the differential equation of the motion by introducing R instead of t as independent variable. Show that for C = 0 the velocity increases linearly with time.
Solution:
| Concepts, principles, relations that apply to the
problem: Newton's 2nd law: F = dp/dt, dp = mdv + vdm |
| Why do they apply? The system consists of the moisture, with zero momentum. and the water droplet with momentum p. In a time interval dt a small amount dm of the moisture receives an impulse vdm. In the same time interval the velocity of the drop changes by dv. The total momentum change of the system is dp = mdv + vdm. |
| How do they apply? Choose a coordinate system such that the z-axis points downward and F = mgk. The problem then is a one-dimensional problem. p = mv, dp/dt = mdv/dt + vdm/dt = mg. dv/dt = g - (v/m)dm/dt. m = ρ4πR3/3, dm/dt = λ4πR2, since the increase in mass per unit time is proportional to the surface area. But we also have dm/dt = ρ4πR2dR/dt. Therefore dR/dt = λ/ρ, R = C + (λ/ρ)t. |
| Details of the calculation: dv/dt = (dv/dR)(dR/dt) = (λ/ρ)(dv/dR). But we also have dv/dt = g - (v/m)(dm/dt) = g - 3vλ/(ρR). Therefore dv/dR = (ρg/λ) - 3v/R. [The differential equation dx/dt + nx/t = B has the general solution x = At-n + Bt/(n+1).] v = AR-3 + (ρg/4λ)R, v = A/(C + (λ/ρ)t)3 + (ρg/(4λ))(C + (λ/ρ)t). At t = 0 v = v0. v0 = A/C3 + ρgC/(4λ), A = v0C3- ρgC4/(4λ). As C goes to zero A goes to zero, and v = (ρg/(4λ))(λ/ρ)t = gt/4. v increases linearly with time. |
Problem 10:
A rocket of initial mass m0 is shot vertically upward. Assume the motion occurs under constant gravitational acceleration g, and that the initial velocity of the rocket on the surface of the earth is zero. The rocket expels 1/100 of its initial mass per second for 50 seconds. The exhaust velocity is 2000 m/s relative to the rocket. What is the maximum height reached by the rocket? Neglect friction.
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 decreases 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 + dmv' (only first order terms are retained). Note: dm is negative. By Newton's 2nd law dp/dt is equal to the external force F = -mg. |
| How do they apply? -mg = m(dv/dt) + (dm/dt)v' Here v' = 2000 m/s and dm/dt = - km0 with k = 1/(100 s). We write dv/dt = -g + km0v'/m. For 0 < t < 50 s we have m = m0 - km0t. Therefore dv/dt = -g + km0v'/(m0 - km0t) = -g + kv'/(1 - kt). |
|
Details of the calculation: Let t' = 1 - kt. Then dv/dt' = g/k - v'/t'. We can integrate to find v(t'1).  v(t1) = -gt1 - v'ln(1 - kt1) At time t1 the height of the rocket is  .h(t1) = -gt2/2 + (v'/k)[(1 -kt1) ln(1 - kt1) + kt1] At t1 = 50 s we have (with g = 9.81 m/s2) v(t1) = -g*50 s - (2000 m/s)ln(0.5) = 896 m/s, h(t1) = -g(50 s)2/2 + (2*105 m)[(0.5)ln(0.5) + 0.5] = 18422 m For t > t1 the rocket is coasting in a gravitational field. It will climb until its speed has decreased to zero. Δh = ½v(t1)2/g = 40918 m. The total height reached is htotal = h(t1) + Δh = 50340 m. |
