3) A ball of mass m is thrown straight up into the air with an initial speed v 0

Question

3) A ball of mass m is thrown straight up into the air with an initial speed v0.(a) Find an expression for the maximum height reached by the ball in terms of v0 and g.

hmax =  

(b) Using conservation of energy and the result of part (a), find the magnitude of the momentum of the ball at one-half its maximum height in terms of m and v0.

p =  

5)A 0.268-kg volleyball approaches a player horizontally with a speed of 12.6 m/s. The player strikes the ball with her fist and causes the ball to move in the opposite direction with a speed of 22.1 m/s.

(a) What impulse is delivered to the ball by the player? (Take the direction of final velocity to be the positive direction. Indicate the direction with the sign of your answer.)
  kg · m/s

(b) If the player's fist is in contact with the ball for 0.0600 s, find the magnitude of the average force exerted on the player's fist.

N

7)A 71-kg fisherman in a 126-kg boat throws a package of mass m = 15 kg horizontally toward the right with a speed of vi = 4.1 m/s as in the figure below. Neglecting water resistance, and assuming the boat is at rest before the package is thrown, find the velocity of the boat after the package is thrown.

(a) What is the speed of the three coupled cars after the collision?
m/s

(b) How much kinetic energy is lost in the collision?
J

a)magnitude m/s b)direction

9) A railroad car of mass 2.85 104 kg moving at 2.60 m/s collides and couples with two coupled railroad cars, each of the same mass as the single car and moving in the same direction at 1.20 m/s.

(a) What is the speed of the three coupled cars after the collision?
m/s

(b) How much kinetic energy is lost in the collision?
J

Explanation / Answer

They want you to fill in a value for v in terms of vi.

At its maximum height, the ball must no longer be moving upwards (or else this wouldn't be the maximum). So the velocity is 0.
p = m*0 = 0

Kinetic energy is 1/2 m v^2
Potential energy is mgh.
The energy at half the max height is 1/2 m v^2 + mg(hmax/2)
The initial energy of the ball is 1/2 m (vi)^2 + 0.
The energy at the max height is 0 + mg(hmax)

All of these must be equal, so:
1/2 m (vi)^2 = 1/2 m v^2 + (1/2 m (vi)^2)/2
v = vi / 2

p = m vi / 2


Alternatively, you can use the equations of ballistic motion.
v = vi - gt
h = vi t - 1/2 g t^2

0 = vi - g(tmax)
tmax = vi / g
hmax = vi (vi / g) - 1/2 g (vi / g)^2 = 1/2 vi^2 / g
hmax / 2 = vi t - 1/2 g t^2

solve for t, then for v.
The result will be the same as the energy method.

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