W1. A steel ball with mass 40.0 g is dropped from a height of 2.00 m onto a
horizontal steel slab. The ball rebounds to a height of 1.60 m. Calculate the
impulse delivered to the ball during impact.
2. In a curling match, a 6.0kg rock with speed 3.50 m/s collides with another
motionless 6.0kg rock. What are the velocities of the rocks after the collision if
it is (a) elastic (b) perfectly inelastic?
3. A 10.0-g marble slides to the left with a velocity of magnitude 0.4m/s on a
frictionless, horizontal surface and has a head-on, elastic collision with a
larger 30.0-g marble sliding to the right with a velocity of 0.2m/s magnitude.
(a) Find the velocity of each marble (magnitude and direction) after the
collision. (b) Calculate the change in momentum.
4. A 200kg load is hung on a wire having a length of 4.00m, cross-sectional
area 0.20x10-4m2, and Young’s modulus 8.00x1010N/m2. What is its
increase in length?
5. A vertical, solid steel post 25 cm in diameter and 2.50m long is required to
support a load of 8000kg. You can ignore the weight of the post. What are
(a) the stress in the post; (b) the strain in the post; and (c) the change in the
post’s length when the load is applied?
6. What is the total mass of the Earth’s atmosphere? (The radius of the Earth is
6.37x106m, and atmospheric pressure at the surface is 1.013x105N/m2.)
7. (a) Calculate the absolute pressure at an ocean depth of 1000m. Assume the
density of seawater is 1024kg/m3 and that the air above exerts a pressure of
101.3kPa. (b) At this depth, what force must the frame around a circular
submarine porthole having a diameter of 30.0cm exert to counterbalance the
force exerted by the water?
8. Calculate the buoyant force, if a steel ball of radius 6 cm is totally immersed
in water.
1. The impulse delivered to the ball during impact can be calculated using the equation:
Impulse = change in momentum
First, calculate the initial velocity of the ball before impact using the equation for gravitational potential energy:
mgh = 1/2 mv^2
(0.04 kg)(9.8 m/s^2)(2 m) = 1/2 (0.04 kg) v^2
v = 6.26 m/s
The impulse can then be calculated as the change in momentum:
Impulse = mΔv
Impulse = (0.04 kg)(6.26 m/s - (-6.26 m/s))
Impulse = 0.04 kg * 12.52 m/s
Impulse = 0.50 N*s
2.
(a) In an elastic collision, both momentum and kinetic energy are conserved. Use the conservation of momentum equation to solve for the final velocities of the rocks.
Initial momentum = final momentum
m1v1i + m2v2i = m1v1f + m2v2f
(6.0 kg)(3.50 m/s) = (6.0 kg)v1f + (6.0 kg)v2f
21 = 6v1f + 6v2f
From conservation of kinetic energy:
(1/2) m1v1i^2 = (1/2) m1v1f^2 + (1/2) m2v2f^2
0 = (6.0 kg)v1f^2 + (6.0 kg)v2f^2 - 21
Solve these two equations simultaneously to find the final velocities of the rocks.
(b) In a perfectly inelastic collision, the two rocks stick together after the collision. Use the conservation of momentum equation to find the final velocity of the combined mass.
m1v1i + m2v2i = (m1 + m2)vf
(6.0 kg)(3.50 m/s) + 0 = (6.0 kg + 6.0 kg)vf
21 = 12vf
vf = 1.75 m/s
3.
(a) First determine the initial total momentum before the collision, and the final total momentum after the collision. Then use conservation of momentum to solve for the velocities of the marbles.
(b) Calculate the change in momentum by finding the difference between the initial and final total momenta.
4. The change in length of the load can be calculated using the equation:
ΔL = (F * L) / (A * Y)
Where F is the force applied (weight of the load), A is the cross-sectional area, Y is the Young's modulus, and L is the original length.
5.
(a) The stress in the post can be calculated using the equation:
Stress = Force / Area
(b) The strain in the post can be calculated using the equation:
Strain = Change in length / Original length
(c) The change in the post's length when the load is applied can be calculated using the stress, strain, and material properties of the post.
6. The total mass of the Earth's atmosphere can be calculated by multiplying the atmospheric pressure at the surface by the surface area of the Earth.
7.
(a) The absolute pressure at an ocean depth of 1000m can be calculated by adding the pressure due to the water depth to the atmospheric pressure above.
(b) The force the frame around the circular submarine porthole must exert to counterbalance the force exerted by the water can be calculated using the pressure difference and the area of the porthole.
8. The buoyant force acting on the steel ball immersed in water can be calculated using Archimedes' principle:
Buoyant force = weight of the water displaced = density of water * volume of the sphere * acceleration due to gravity.