Problem#1
A 3.00-kg particle has a velocity of (3.00i – 4.00j) m/s. (a) Find its x and y components of momentum. (b) Find the magnitude and direction of its momentum.
Answer:
Given: m = 3.00 kg and v = (3.00i – 4.00j) m/s,
(a) x and y components of momentum is
p = mv = (3.00 kg)(3.00i – 4.00j) m/s
p = (9.00i – 12.0j) m/s
then,
px = 9.00 kgm/s and py = –12.0 kgm/s
(b) the magnitude of its momentum is
p = [(px)2 + (py)2]1/2 = [(9.00 kgm/s)2 + (–12.0 kgm/s)2]1/2
p = 15.0 kgm/s
and the direction of its momentum is
θ = tan-1(py/px) = tan-1(–12.0/9.00) = 3070.
Problem#2
A 0.100-kg ball is thrown straight up into the air with an initial speed of 15.0 m/s. Find the momentum of the ball (a) at its maximum height and (b) halfway up to its maximum height.
Answer:
(a) the momentum of the ball at its maximum height is p = 0, because at maximum height v = 0.
(b) Its original kinetic energy is its constant total energy,
Ki = ½ mvi2 = ½ (0.100 kg)(15.0 m/s)2 = 11.2 J
At the top all of this energy is gravitational. Halfway up, one-half of it is gravitational and
the other half is kinetic:
½ Ki = ½ mv2
11.2 J = (0.100 kg)v2
v = 10.6 m/s
then p = mv = (0.100 kg)(10.6 m/s j) = 1.06j kgm/s
Problem#3
How fast can you set the Earth moving? In particular, when you jump straight up as high as you can, what is the order of magnitude of the maximum recoil speed that you give to the Earth? Model the Earth as a perfectly solid object. In your solution, state the physical quantities you take as data and the values you measure or estimate for them.
Answer:
I have mass 85.0 kg and can jump to raise my center of gravity 25.0 cm. I leave the ground with
speed given by
vf2 = vi2 + 2a∆x
0 = vi2 + 2(–9.80 m/s2)(0.250 m)
vi = 2.20 m/s
Total momentum of the system of the Earth and me is conserved as I push the earth down and
myself up:
pi = pf
0 = MEvE + mv
0 = (5.98 x 1024 kg)vE + (85.0 kg)(2.20 m/s)
vE = 3.13 x 10-23 kgm/s
Problem#4
Two blocks of masses M and 3M are placed on a horizontal, frictionless surface. A light spring is attached to one of them, and the blocks are pushed together with the spring between them (Fig. 1). A cord initially holding the blocks together is burned; after this, the block of mass 3M moves to the right with a speed of 2.00 m/s. (a) What is the speed of the block of mass M ? (b) Find
the original elastic potential energy in the spring if M = 0.350 kg.
Answer:
For the system of two blocks ∆p = 0,
or pi = pf
therefore, 0 = Mv + 3Mu
0 = Mv + 3M(2.00 m/s)
v = –6.00 m/s
(b) the original elastic potential energy in the spring if M = 0.350 kg is
Us = ½ kx2 = ½ Mv2 + ½ (3M)u2
Us = ½ (0.350 kg)(–6.00 m/s)2 + ½ 3(0.350 kg)(2.00 m/s)2
Us = 8.40 J
Problem#5
(a) A particle of mass m moves with momentum p. Show that the kinetic energy of the particle is K = p2/2m. (b) Express the magnitude of the particle’s momentum in terms of its kinetic energy and mass.
Answer:
(a) The momentum is p = mv, so v = p/m and the kinetic energy is
K = ½ mv2 = ½ m(p/m)2
K = ½ p2/m
(b) from (a) we get p = (2mK)1/2
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