Physics 101 - Homework # 6 Solutions

Note: Symbols written in Bold are vectors.

Chapter 8

Problem 28)

The ²⁴¹Am nucleus is at rest before emitting the particle, so the particle and ²³⁷Np nucleus must move in opposite directions, which makes this a one-dimensional situation.
a) Because momentum is conserved, we can write: m_Amv_Am = mv + m_Npv_Np; 0 = (4m₀)v + (237m₀)v_Np
which gives: v_Np = -(4/237)v
The released energy (=Q) must be the kinetic energies of the emitted products (subtle point: this assumes that the particles are emitted in their ground states rather than in excited states):: Q = ½mv² + ½m_Npv_Np²; 9.6 x 10^-13 J = ½(4m₀)v² + ½(237m₀)(4/237)²v²; = 2.034m₀v²
where m₀ = 1.66 x 10^-27 kg: v = 1.70 x 10⁷ m/s; v_Np = 2.85 x 10⁵ m/s
b) The kinetic energies are: K = ½(4)(1.66 x 10^-27 kg)(1.70 x 10⁷ m/s²)² = 9.58 x 10^-13 J; K_Np = ½(237)(1.66 x 10^-27 kg)(2.85 x 10⁵ m/s²)² = 1.59 x 10^-14 J
Almost all the energy is carried off by the lighter product (the particle).

Problem 32)

a) Because momentum is conserved, we can write: m₁v₁ + m₂v₂ = m₁v₃ + m₂v₄; (0.30 kg)(1.4 m/s)i + (0.15 kg)(-2.5 m/s)i = (0.30 kg)v₃i + (0.15 kg)v₄i
Because the collision is elastic, the relative velocity does not change:: v₁ - v₂ = - (v₃ - v₄)
or: (1.4 m/s) i - [ - (2.5 m/s)i] = - (v₃i - v₄i)
Combining these two equations, we get, after a bit of algebra,: v₃ = - 1.2 m/s and v₄ = 2.7 m/s
The total initial kinetic energy is: K_i = ½m₁v₁² + ½m₂v₂²; = ½(0.30 kg)(1.4 m/s)² + ½(0.15 kg)(-2.5 m/s)² = 0.76 J

Problem 43)

For the collision of the two masses, we use momentum conservation:: mv + 0 = mv₁ + mv₂; (2.50 m/s)i = [(0.50 m/s)i + ( -1.00 m/s)j] + v₂
which gives: v₂ = (2.00 m/s)i + (1.00 m/s)j
To test whether the collision is elastic, we find the kinetic energy before and after:: K_i = ½m(2.50 m/s)² = (3.13 m) J; K_f = ½m[ (0.50 m/s)² + (-1.00 m/s)² + (2.00 m/s)² + (-1.00 m/s)² ] = (3.13 m) J
These are equal no matter what m is. Because the kinetic energy is conserved, the collision is elastic.
Since the masses are the same, we could also show that the final velocities are perpendicular: v₁·v₂ = [(0.50 m/s)i + (-1.00 m/s)j]·[(2.00 m/s)i + (1.00 m/s)j] = 0
since that is also true only ff the collision was elastic (see text, Eq 8-44).

Problem 49)

We use momentum conservation for the collision; consider the x and y-components separately:
x-direction:: m_carv_car,i + 0 = m_carv_car,fcos_car + m_truckv_truck,fcos_truck; (1000 kg)v_car,i = (1000 kg)v_car,fcos60° + (1500 kg)(21.6 m/s)cos33.7°
y-direction:: 0 + 0 = -m_carv_car,fsin_car + m_truckv_truck,fsin_truck; 0 = -(1000 kg)v_car,fsin60° + (1500 kg)(21.6 m/s)sin33.7°
These are two equations in two unknowns; solving, we get: v_car,f = 20.8 m/s and v_car,i = 37.4 m/s = 84 mi/h.
The driver of the car was speeding (unless they were driving down the Autobahn...).

Problem 52)

We find the center of mass from (the sums run over i): R = mr_i/m_i; = [(0.2 kg)0 + (0.1 kg)(0.3 m)j + (0.3 kg){(0.2 m)i + (0.2 m)j}]/(0.2 kg + 0.1 kg + 0.3 kg); = (0.10 m)i + (0.15 m)j

Problem 57)

We ignore the size of the iron block, and treat it as a point mass. The center of mass of the handle will be at its midpoint, from symmetry arguments. We choose the origin at the bottom of the handle. For the system, we have: X = m_ix_i/m_i; = [(4.0 kg)(0) + (1.8 kg)(0)]/(4.0 kg + 1.8 kg) = 0 m; Y = m_iy_i/m_i; = [(4.0 kg)(1.2 m) + (1.8 kg)(0.6 m)]/(4.0 kg + 1.8 kg) = 1.0 m from the bottom of the handle.

Problem 73)

We choose the positive direction to be the original direction of motion, which we take to be to the right. The center-of-mass velocity is found from (see Eq. 8-50 in the text, for example):
a) v_CM = (m₁v₁ + m₂v₂)/(m₁ + m₂): = [(48 kg)(6 m/s) + (82 kg)(9 m/s)]/(48 kg + 82 kg) = 7.9 m/s (to the right)
b) Because there are no outside (or external) forces, the velocity of the center of mass will not change:: v_CM = 7.9 m/s (to the right)
We can show this explictly by calculating the new velocity of the pair after they collide using conservation of momentum, and then calculating v_CM similarly to what was done in part a).
c) The center of mass frame is a reference frame moving with the velocity of the center-of-mass; in this reference frame, we have: v₁' = v₁ - v_CM = 6 m/s - 7.9 m/s = - 1.9 m/s (to the left); v₂' = v₂ - v_CM = 9 m/s - 7.9 m/s = + 1.1 m/s (to the right)
d) We find the average force on the skaters from the momentum change (the impulse):: J = F_avt = p; F_av = p/t = (48 kg + 82 kg)(0 - 7.9 m/s)/(0.05 s) = -2.1 x 10⁴ (to the left)
Therefore, the force on the barrier is the reaction to this force:: 2.1 x 10⁴ N (to the right)
The momentum of the two-skater system is not conserved any more since in the collision with the barrier there is now a net external force (caused by the barrier).

Problem 75)

a)
b) In the horizontal direction, momentum is conserved:: 0 = p_proj cos35° - p_cannon; 0 = (5 kg)(800 m/s)cos35° - (800 kg)v_cannon
which gives the cannon's recoil velocity as: v_cannon = 4.1 m/s
along the ground.
c) The recoiling cannon's component of momentum perpendicular to the ground remains zero because of the upward impulse provided by the ground; the earth therefore recoils (infinitesimally) downwards, conserving the total vertical component of momentum.

Problem 78)

We choose the boxcar and the water as a system, with the origin at the end of the boxcar with the tank. Because there are no horizontal external forces, the horizontal momentum will stay=zero always, and the center of mass of the system will not move horizontally. We can find the center of mass, before the tank leaks, from: X = m_ix_i/ m_i; = {2400 kg)(0.5 m) + (3200 kg)(4 m)]/[2400 kg + 3200 kg] = 2.5 m
After the tank empties, mass is distributed uniformly, so the center of mass must be X' = 4.0 m from the end. Because the center of mass does not move, the boxcar must move X'-X = 1.5 m toward the end where the water was.