A giant "egg" explodes as part of a fireworks display. The egg isat rest before the explosion, and after the explosion, it breaksinto two pieces, with the masses indicated in the diagram,traveling in opposite directions. Part A mass = 10kg. Part B mass=30kg.What is the momentum p_A,i of piece A before the explosion?The momentum of piece B is measured to be 500 kg *m/s after theexplosion. Find the momentum p_A,f of piece A after the explosion.

OneClass: A giant "egg" explodes as part of a fireworks display. The egg isat rest before the explosi...

A playground ride consists of a disk of mass M = 56 kg and radius R= 2.0 m mounted on a low-friction axle. A child of mass m = 21 kgruns at speed v = 2.8 m/s on a line tangential to the disk andjumps onto the outer edge of the disk.

ANGULAR MOMENTUM
(a) Consider the system consisting of the child and the disk, butnot including the axle. Which of the following statements are true,from just before to just after the collision?
The momentum of the system doesn't change.
The momentum of the system changes.
The axle exerts a force on the system but nearly zero torque.
The torque exerted by the axle is zero because the force exerted bythe axle is very small.
The angular momentum of the system about the axle changes.
The angular momentum of the system about the axle hardlychanges.
The torque exerted by the axle is nearly zero even though the forceis large, because || is nearly zero.

(b) Relative to the axle, what was the magnitude of the angularmomentum of the child before the collision?
|C| = kgÂ·m2/s
(c) Relative to the axle, what was the angular momentum of thesystem of child plus disk just after the collision?
|C| = kgÂ·m2/s
(d) If the disk was initially at rest, now how fast is it rotating?That is, what is its angular speed? (The moment of inertia of auniform disk is Â½MR2.)
= radians/s
(e) How long does it take for the disk to go around once?
Time to go around once = s
ENERGY
(f) If you were to do a lot of algebra to calculate the kineticenergies before and after the collision, you would find that thetotal kinetic energy just after the collision is less than thetotal kinetic energy just before the collision. Where has most ofthis energy gone?
Increased chemical energy in the child.
Increased thermal energy of the disk and child.
Increased translational kinetic energy of the disk.

MOMENTUM
(g) What was the speed of the child just after the collision?
v = m/s
(h) What was the speed of the center of mass of the disk just afterthe collision?
vcm = m/s
(i) What was the magnitude of the linear momentum of the disk justafter the collision?
|p| = kgÂ·m/s
(j) Calculate the change in linear momentum of the systemconsisting of the child plus the disk (but not including the axle),from just before to just after impact, due to the impulse appliedby the axle. Take the x axis to be in the direction of the initialvelocity of the child.
px = px,f - px,i = kgÂ·m/s

ANGULAR MOMENTUM
(k) The child on the disk walks inward on the disk and ends upstanding at a new location a distance R/2 = 1 m from the axle. Nowwhat is the angular speed? (It helps to do this analysisalgebraically and plug in numbers at the end.)
= radians/s
ENERGY
(l) If you were to do a lot of algebra to calculate the kineticenergies in part (k), you would find that the total kinetic energyafter the move is greater than the total kinetic energy before themove. Where has this energy come from?
Decreased chemical energy in the child.
Decreased momentum of the disk.
Decreased thermal energy of the disk and child.

Here is a popular lecture demonstration that you can perform athome. Place a golf ball on top of a basketball, and drop the pairfrom rest so they fall to the ground. (For reasons that shouldbecome clear once you solve this problem, do not attempt to do thisexperiment inside, but outdoors instead!) With a little practice,you can achieve the situation pictured here: The golf ball stays ontop of the basketball until the basketball hits the floor. The massof the golf ball is 0.0459 kg, and the mass of the basketball is0.591 kg.

a) If the balls are released from a height where the bottom of thebasketball is at 0.741 m above the ground, what is the absolutevalue of the basketballâs momentum just before it hits theground?

b) What is the absolute value of the momentum of the golf ball atthis instant?

c) Treat the collision of the basketball with the floor and thecollision of the golf ball with the basketball as totally elasticcollisions in one dimension. What is the absolute magnitude of themomentum of the golf ball after these collisions?

d) Now comes the interesting question: How high, measured from theground, will the golf ball bounce up after its collision with thebasketball? (Hint: do not forget to add the diameter of thebasketball, 24.15 cm).

http://www.flickr.com/photos/60996491@N03/5592819537/

i found a and b section a)2.253 kg*m/s and b)0.175 kg*m/s but idont know c and d. please help me....

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