Answers - Center of Mass, Momentum, Collisions

1.

Impulse J = p_f - p_i = mv_f - mv_i. Taking the return direction as positive,
J = 0.060 kg{(40 - (-50)}m/s = 5.4 kg-m/s.

2.

X_CM = (m₁x₁ + m₂x₂ + m₃x₃ + m₄x₄)/( m₁ + m₂ + m₃ + m₄)
       = {(1.0 kg)(0) + (4.0 kg)(2.0 m) + (2.0 kg)(0) + (3.0 kg)(2.0 m)}/(10.0 kg)
       = 1.4 m

Y_CM = (m₁y₁ + m₂y₂ + m₃y₃ + m₄y₄)/( m₁ + m₂ + m₃ + m₄)
       = {(1.0 kg)(0) + (4.0 kg)(0) + (2.0 kg)(2.0 m) + (3.0 kg)(2.0 m)}/(10.0 kg)
       = 1.0 m

3.

Taking east in the +i direction and north in the +j direction,

(a) J = p_f - p_i = (1500 kg)(8.0 m/s)(j - i) = 12 x 10³ kg-m/s(j - i)

(b) J = F_av t F_av = J/t = {12 x 10³ kg-m/s(j - i)}/3.0 s =4 x 10³ N(j - i)

(c) By Newton's third law of motion, the average force of the car on the road
     = -4 x 10³ N(j - i)

4.

Divide the system into three pairs of particles, consisting of 1 + 2,  3 + 4, and
5 + 6. The midway for 1 + 2 is at a and the midpoint for 5 + 6 is c. We can treat 1 + 2 as a point particle of mass of 2m at a and 5 + 6 as a point particle of mass 2m at c. We can combine these two point particles to be another point particle with mass 4m at b. 3 + 4 have their midpoint at b so we can replace them with a point particle of mass 2m at b. Finally the whole works acts like a point particle of mass 6m at b.

5.

(a) p_A = (1.0 kg)(4.0 m/s) = 4.0 kg-m/s
     p_B = (2.0 kg)(2.0 m/s) = 4.0 kg-m/s

(b) K_A = 1/2(1.0 kg)(4.0 m/s)² = 8.0 J
     K_B = 1/2(2.0 kg)(2.0 m/s)² = 4.0 J

While the momenta are equal, the kinetic energies are not.

6.

Take the distance to the edge of the rink as x. Then the velocity v₁ after the collision for the skater of mass m₁ equals +x/12 s and the velocity v₂ of the other skater equals -x/18 s. From conservation of momentum, p_i = p_f,
0 = (50 kg)(x/12 s) + (m₂)(-x/18 s), or (m₂)(x/18 s) = (50 kg)(x/12 s).
m₂ = (50 kg)((18/12) = 75 kg.

7.

Take to the right to be positive. From conservation of momentum,

8.

The center of mass of the four particles at the base of the pyramid is the center of the base. The center of mass of the five-particle system lies on the line perpendicular to the base and passes through the peak of the pyramid. This line passes through the center of mass of the four particles at the base and gives the x-coordinate of the center of mass at the center of the base.
Y_CM = {(m)(h) + (4m)(0)}/5m = h/5.

9.

In the figure above, b represents the center of mass of the barge and x the distance the barge moves to the right as the man moves to the left. No external forces act on the barge-man system and the center of mass CM of this system remains at rest.

Initially,

    X_CM = {(75 kg)(10 m) + (200 kg)(b)}/(75 + 200)kg        (Equation 1)

Finally,

    X_CM = {(75 kg)(8 m) + (200 kg)(b + x)}/(75 + 200)kg    (Equation 2)

Since the CM of the system remains at rest, equate Eq. 1 and Eq. 2 and cancel denominators:

10.

Momentum is conserved:

Since the final momentum of product 1, p_1f, = -p_2f,  the negative of the final momentum of product 2, the initial momentum p_in of the nucleus must equal zero.

11.

12.

Another conservation law is at work here. In an elastic collision, kinetic energy is also conserved. For an elastic collision, (b) cannot occur.

13.

Conservation of momentum:

After the collision, the blocks move to the left with a speed of 1 m/s.

14.

For an elastic collision,

where I have assumed both final velocities to the right. In any calculation, if one of them is to the left, you will get a minus answer in your solution.

Solving Eq. 1 and Eq. 2 simultaneously for v_1f and v_2f gives:

v_1f = {(m₁ - m₂)/(m₁ + m₂)} v_1i, and
v_2f = {2m₁/(m₁ + m₂)} v_1i

For v_1i = 2.0 m/s, m₁ = 1.0 kg, and

(a) m₂ = 1.0 kg, v_1f = 0 and v_2f = 2.0 m/s = v_1f.
Example: cue ball colliding with another billiard ball.

(b) m₂ = 1000 kg, v_1f = -(999/1001)(2.0 m/s) ≈(1)(2.0 m/s) = -v_1i.
v_2f = (2.0/1001)(2.0 m/s) ≈ 0.004 m/s, which is small compared to v_1i.
A basketball colliding with a bowling ball is an example of a relatively small mass colliding with a large mass.

(c) m₂ = 0.001 kg, v_1f = (0.999)(2.0 m/s) ≈ (1)(2.0 m/s) = v_1i.
v_2f = (2.0/1.001)(1.0 m/s) ≈ 2 v_1i. A bowling ball colliding with a basketball is an example of a relatively large mass colliding with a small mass.

15.

If momentum, a vector quantity is conserved, its components are conserved.

16.

17.

Momentum is conserved.

So from Eq. 1,  v_1i = 2 (v_1f)_x = 2v_1f cos Q₁.   (Equation 3)

From conservation of kinetic energy,

From Eq. (3),

v_1i = 2 (v_1i/√2) cos Q₁
cos Q₁ = √2/2, and
Q₁ = 45⁰
Q₂ = -45⁰

18.

(a) Take the zero of gravitational potential energy at the bottom of the ramp, U_af = 0. At the top of the ramp, U_ai = mgh, where h = 0.40 m. On the ramp, from conservation of energy,

(b) When A collides with B, momentum is conserved.

pbi	=	pbf
mAvib	=	(mA + mB)vfb
vfb = [mA/(mA + mB)]vib = [2.0/(11.2)]2.8 m/s = 0.50 m/s

(c) Work done by friction = (U_cf + K_cf ) - (U_ci + K_ci)
                     fd cos 180⁰ = (0 + 0) - [0 - 1/2(11.2 kg)(0.50 m/s)²]
                         -(2.8 N)d = -1.4 N-m   and   d = 0.50 m.

19.

20.

It takes 0.80 N to keep the 5.0 kg moving with constant velocity. The increase in the normal force due to the weight of the bullet does not change the frictional force to two significant figures. Take the gravitational potential at the surface = 0 = U_fb = U_ib.

21.

Take the potential energy at the bottom of swing to be zero. The kinetic energy at the top of the swing is zero since the velocity of A there v_bi = 0. See the figure for #21, above, for parts (a), (b), (c) and (d).

(a) U_bf + K_bf = U_bi + K_bi
         0 + K_bf = (2.0kg)(10 m/s²)(0.80 m) + 0
               K_bf = 16 J = 1/2(2.0 kg)v_bf².

(b) v_bf = 4.0 m/s

(c) p_i = p_f 2.0 kg(4.0 m/s) = (2.0 + 3.0)kg v_cf.  v_cf = 1.6m/s.

(d)                   U_df + K_df = U_di + K_di
     5.0 kg(10 m/s²)h + 0 = 0 + 1/2(5.0 kg)(1.6m/s)²
                                    h = 0.13 m

22.

(a) Do the problem in steps. First there is the collision between the bullet of mass m = 5.0 x 10^-3 kg travelling with a speed of 400 m/s and the block of mass M = 1.0 kg. After the collision, the bullet has a velocity v and the block a velocity v_M. From conservation of momentum,

Just after the collision and just before the block compresses the spring, the initial potential energy of the block-spring system is 0 and the kinetic energy of the block is 1/2 (1.0 kg) v_M². When the spring has been compressed to its maximum, the block momentarily comes to rest and its "final" kinetic energy is zero, but the potential energy of the block-spring system is 1/2 kx² =
1/2(900 N/m)((5 x 10^-2m)². From conservation of energy,

Substituting this value of v_M back into Eq. 1:
   (5.0 x 10^-3 kg)(400m/s) = (1.0kg)(1.5m/s) + (5.0 x10^-3 kg)v or v = 100 m/s.

(b) For the collision,

	Ki = 1/2 (5 x 10-3 kg)(400 m/s)2 = 400 J
	Kf = 1/2(5.0 x 10-3 kg)(100 m/s)2 + 1/2(1.0 kg)(1.5 m/s)2 = (25 + 1.12)J
	K lost = (400 - 26.12)J = 374 J

23.

This problem is a reminder to choose your system carefully. When you drop an object, its momentum does not stay constant because the gravitational attraction of the earth, an external force, acts on it. If you take your system to be the object and the earth, then momentum is conserved. Let the object be a textbook with mass m_b = 2.0 kg and velocity v_b. The earth has a mass M_e = 5.98 x 10²⁴ kg. Before you release the book, the momentum of the system is zero.

p_i = p_f
0 = m_bv_b + M_ev_e, where v_e = the speed acquired by the earth
v_e = (m_b/M_e)v_b = (2.0 kg/5.98 x 10²⁴ kg)v_b = 3.3 x 10^-25 v_b

So the earth does move up, but we do not detect the motion.