Problem 1:
The mass of the blue puck is 20% greater than the mass of the green one. Before colliding, the pucks approach each other with equal and opposite momenta, and the green puck has an initial speed of 10 m/s. Find the speed of the pucks after the collision, if half the kinetic energy is lost during the collision.
Solution:
 | Concepts, principles, relations
that apply to the problem: Inelastic collisions, momentum conservation |
| Why do they apply? The collision is inelastic, since energy is not conserved. |
| How do they apply? The total momentum of the two pucks is zero before the collision and after the collision. Let particle 1 be the green puck and particle 2 be the blue puck. Before and after the collision the ratio of the speeds is v2/v1 = m1/m2 = 1/1.2. |
| Details of the calculation: The final kinetic energy of the system equals 1/2 times its initial kinetic energy. (1/2)m1v1i2 + (1/2)m2v2i2 = 2((1/2)m1v1f2 + (1/2)m2v2f2). m1v1i2 + m2v2i2 = 2(m1v1f2+m2v2f2). m1v1i2 + 1.2m1(v1i/1.2)2 = 2(m1v1f2 + 1.2m1(v1f/1.2)2). m1(v1i2 + v1i2/1.2) = 2m1(v1f2 + v1f2/1.2). v1i2 = 2v1f2, v1f = 0.707v1i. v1f = 7.07 m/s, v2f = 5.89 m/s. |
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 particle of mass m strikes a frictionless, smooth, hard surface at an angle q from the surface normal. The particle bounces off with coefficient of restitution e. (The coefficient of restitution is defined as the ratio of the velocity components along the normal to the plane of contact after and before the collision.) Find the rebound angle for the particle as it leaves the surface.
Solution:
| Concepts, principles, relations
that apply to the problem: Definition: coefficient of restitution |
| Why do they apply? This is a perfectly inelastic collision. We are asked to find the rebound angle when given the coefficient of restitution. |
| How do they apply? e = |v2|/|v1|, v1 is the velocity component of the particle toward the floor before collision and v2 is the velocity component of the particle away from the floor after the collision. Before the collision: tanq = |vx|/|vy|. After the collision: tanq' = |vx|/|vy| = |vx|/(e|vy|) = (1/e)tanq. Since e < 1, the rebound angle is bigger than the incoming angle. |
| Details of the calculation: None |
Problem 4:
A uniform heavy chain of mass
l
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 ldxv to zero in a time interval dt. |
| How do they apply? The table has to exert a force F = lxg + lv2 in the upward direction. F = lxg + l2gx = 3lxg. |
| Details of the calculation: None |
Problem 5:
A rocket of initial mass m0 ejects fuel at a constant rate km0 and at a velocity v' relative to the rocket shell of mass m1. Show that the minimum rate of fuel consumption that will allow the rocket to rise at once is k = g/v' where g is the gravitational acceleration. Find the greatest speed achieved and the greatest height reached under that condition.
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 - |dm|v' (only first order terms are retained). 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 |dm/dt| = km0. Therefore dv/dt = -g + km0v'/m. At t = 0 m = m0. We need dv/dt ³ 0 for the rocket to get off the ground, therefore we need k ³ g/v'. For t > 0 and m > m1 we have m = m0 - km0t. Therefore dv/dt = -g + km0v'/(m0 - km0t) = -g + kv'/(1 - kt). At time tc the fuel is used up. m1 = m0 - km0tc, tc = (1 - m1/m0)/k = (1 - m1/m0)v'/g under the conditions stated in the problem. |
| Details of the calculation: For t < tc, dv/dt = -g + kv'/(1 - kt). Let t' = 1 - kt. Then dv/dt' = g/k - v'/t'. We can integrate to find v(t').  t1' = 1 - kt1, t0' = 1. v(t1) = -gt1 - v'ln(1 - kt1) = -gt1 - v'ln(1 - gt1/v'). At t1 = tc, v(tc) = -v'(1 - m1/m0) - v' ln(m1/m0). This is the greatest speed of the rocket. At time tc the height of the rocket is  . .For t > tc the rocket is coasting in a gravitational field. It will climb until its speed has decreased to zero. Dh = ½v(tc)2/g. The total height reached is htotal = h(tc) + Dh. |