1 Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by

Question

1 Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, they push off against one another. Adolf has a mass of 100 kg, and Ed has a mass of 86 kg. Following the push, Adolf swings upward to a height of 0.58 m above his starting point. To what height above his own starting point does Ed rise?

2 Batman (mass = 97.3 kg) jumps straight down from a bridge into a boat (mass = 672 kg) in which a criminal is fleeing. The velocity of the boat is initially +11.0 m/s. What is the velocity of the boat after Batman lands in it?

3 A 1.80-kg ball, moving to the right at a velocity of +3.09 m/s on a frictionless table, collides head-on with a stationary 8.20-kg ball. Find the final velocities of (a) the 1.80-kg ball and of (b) the 8.20-kg ball if the collision is elastic. (c) Find the magnitude and direction of the final velocity of the two balls if the collision is completely inelastic.

4 A stream of water strikes a stationary turbine blade horizontally, as the drawing illustrates. The incident water stream has a velocity of +16.0 m/s, while the exiting water stream has a velocity of -16.0 m/s. The mass of water per second that strikes the blade is 32.0 kg/s. Find the magnitude of the average force exerted on the water by the blade.

Explanation / Answer

1) The height is inversely proportional to the mass

h2/h1 = m1/m2
h2 =0.58 *100/ 86] = 0.674 m

2)Momentum after = momentum before

Mo before = mass * velocity = 672 kg * 11 m/s = 7392 kg-m/s

7392 kg-m/s = (672 + 97.3) * v


v = 9.998 m/s

3) elastic collision:

v1' = 2(m1v1 + m2v2)/(m1+m2) - v1
v1' = 2(4.6*1.86 + 9.9*0)/(14.5) - 1.86 = - 0.6798 m/s <--- the first ball bounces back with 0.6798 m/s

with v1' = velocity of the first ball after collision, v1, m1 = velocity of first ball before collision,
m2, v2 = velocity and mass of second ball before collision.

v2' = 2(m1v1 + m2v2)/(m1+m2) - v2
v2' = 2(4.6*1.86 + 9.9*0)/(14.5) - 0
v2' = 1.18 m/s <--- velocity of second ball after collision (in direction of first ball before collision)
----------
inelastic collision:

v' = (v1 m1 + v2m2) / (m1+m2)
v' = (4.6*1.86)/(14.5) = 0.59 m/s <-- velocity of ball 1 and 2 sticking together

4). Each second the impulse, or momentum change, is m?V = 32 kg * 32 m/s. Impulse has the units mass*velocity or force*time; thus an impulse of 1024 N-s is added each second. Divide by time and you have F = 1024 N

Related Questions

Navigate

• Browse (All)
• Browse 1
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.