PHY101 - Mastering Physics #6
A child swings back and forth on a swing suspended by 2.5 m -long ropes. Find the turning-point angles if the child has a speed of 0.90 m/s when the ropes are vertical. Express your answer using two significant figures.
(1/2)mVmax^2 = mgl(1 - cos?) 1 - cosθ = Vmax^2/(2gl) 1 - cosθ = (0.9)^2/(2*9.81*2.5) 1 - cosθ = 0.01 cosθ = 0.983 θ = 10
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m2, rises through a height h. (Figure 1) A) Find the speed of the masses just before m2 lands, giving your answer in terms of m1, m2, g, and h. Assume the ropes and pulley have negligible mass and that friction can be ignored. Express your answer in terms of the variables m1, m2, h, and appropriate constants.
A) Acceleration produced when the machine starts to move: a = [(m2 -m1)/(m1 + m2)]g Distance traveled is s = h Initial velocity: vi = 0 m/s vf^2 - vi^2 = 2as vf^2 - 0^2 = 2[(m2 -m1)/(m1 + m2)]gh vf = √(2gh * [(m2 -m1)/(m1 + m2)])
A plate drops onto a smooth floor and shatters into three pieces of equal mass. Two of the pieces go off with equal speeds v at right angles to one another. Find the speed of the third piece. Express your answer in terms of the variable v. Find the direction of the third piece. Assume the motion of the two pieces at right angles to one another is in the positive x and y directions.
A) Law of Conservation of Momentum: p1 + p2 + p3 = 0 (mvx * x) + (mvy * y) + mv3 = 0 mv3 = -[(mvx * x) + (mvy * y)]/m = - (vx + vy) I v3 I = √((-v)^2 + (-v)^2) v3 = 2√v B) θ = tan-1(y/x) θ = 180 - tan-1(-v/-v) θ = 180 - 45 = 135
Force A has a magnitude F and acts for the time Δt, force B has a magnitude 2F and acts for the time Δt/3, force C has a magnitude 5F and acts for the time Δt/10, and force D has a magnitude 10F and acts for the time Δt/100. Rank these forces in order of decreasing impulse. Rank from largest to smallest. To rank items as equivalent, overlap them.
A>B>C>D Impulse = F x time IA = F*deltat IB = 2F * deltat./3 = 0.67 F deltat IC = 5F * deltat/10 = 0.5 F deltat ID = 10F * deltat/100 = 0.10 F deltat
A 0.24 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0. B) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 3.0 m. Express your answers using two significant figures separated by commas.
B) U = (0.24 kg)(9.81 m/s^2)(3.0 m) U = 7.0 J K = E - U K = 9.4J - 7.0J K = 2.4 J
A skateboarder starts at point A in the figure(Figure 1) and rises to a height of 2.64 m above the top of the ramp at point B. What was the skateboarder's initial speed at point A?
KA + UA = KB + UB (1/2)mv2 + mgh = 0 + mg(h + 2.64m) (1/2)mv2+ mgh = mgh + mg(2.64m) v = √(sg2.64) = 7.19 m/s
A 0.130-kg baseball is dropped from rest. If the magnitude of the baseball's momentum is 1.70 kg⋅m/s just before it lands on the ground, from what height was it dropped? Express your answer using three significant figures.
Momentum= m*vf 1.70 = 0.130 *vf vf = 13.1 m/s vf^2 = v^2 +2*g(h (13.1)^2 = 0^2 +2*9.8*h h = 8.72
Starting at rest at the edge of a swimming pool, a 72.0 kg athlete swims along the surface of the water and reaches a speed of 1.20 m/s by doing the work Wnc1 = 171 J . Find the nonconservative work, Wnc2, done by the water on the athlete.
Wc + Wnc = total energy = 1/2Mv^2 Total Energy = 1/2Mv^2 = 51.8 171J + Wnc = 51.8 J Wnc = -119 J
A 95-kg astronaut and a 1000-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, giving it a speed of 0.25 m/s directly away from the shuttle. Seven-and-a-half seconds later the astronaut comes into contact with the shuttle. What was the initial distance from the shuttle to the astronaut? Express your answer using two significant figures.
m1 = 95 kg m2 = 1000 kg v2 = 0.25 m/s t = 7.5 s From conservation of momentum: m1v1 = m2v2 or v1 = (m2/v2)m1 rom the equation of motion starting at rest with no acceleration: x = v1t = (m2/v2)m1t x = 19.7 m
A 17.5 kg dog is running northward at 2.72 m/s , while a 5.57 kg cat is running eastward at 3.68 m/s . Their 79.2 kg owner has the same momentum as the two pets taken together. A) Find the direction and magnitude of the owner's velocity.
A) Mass of dog is m1 = 17.5 kg Velocity of dog is u1 = 2.72 m/s (north) Mass of cat is m2 = 5.57 kg Velocity of cat is u2 = 3.68 m/s (east ) Mass of owner is M = 79.2 kg Momentum of dog is Py = m1 x u1 Py = 47.6 J Momentum of cat is Px = = m2 x u2 Px = 20.5 Net magnitude of momentum is: P = (Px2 + Py2)^1/2 P = 51.8 J Angle with the +x axis: θ =tan-1(Py / Px ) θ = 66.7 B) Velocity of owner is v = P / M v = 0.65
A 0.24 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0. A) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 4.0 m. Express your answers using two significant figures separated by commas.
A) Since velocity at the top is 0, the kinetic energy of the apple at the top is zero (K = 1/2mv^2) K = 0 Gravitational Potential Energy (U): U = (0.24 kg)(9.81 m/s^2)(4.0 m) U = 9.4 J Total mechanical energy at 4.0m: E = 0J + 9.4J = 9.4J
The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, the large mass, m2, falls through a height h and hits the floor, and the small mass, m2, rises through a height h. (Figure 1) B) Evaluate your answer to part A for the case h = 1.0 m , m1 = 3.8 kg , and m2 = 4.5 kg . Express your answer using two significant figures.
B) m1 = 3.8 kg m2 = 4.5 kg h = 1.0 m g = 9.81 m/s^2 vf = √(2(9.81)(1.0) * [(4.5 -3.8)/(3.8 + 4.5)]) vf = 1.3 m/s
A 0.24 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0. C) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 2.0 m. Express your answers using two significant figures separated by commas.
C) U = (0.24 kg)(9.81 m/s^2)(2.0 m) U = 4.7 J K = 9.4J - 4.7J K = 4.7 J
A 0.24 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0. D) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 1.0 m. Express your answers using two significant figures separated by commas.
D) C) U = (0.24 kg)(9.81 m/s^2)(1.0 m) U = 2.4 J K = 9.4 J - 2.4 J K = 7.0 J
A 0.24 kg apple falls from a tree to the ground, 4.0 m below. Ignore air resistance. Take ground level to be y=0. E) Determine the apple's kinetic energy, K, the gravitational potential energy of the system, U, and the total mechanical energy of the system, E, when the apple's height above the ground is 0 m. Express your answers using two significant figures separated by commas.
E) U = 0 E = 9.4 J K = 9.4 J
