A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.500 m/s on the frictionless, horizontal surface of an icy, New York sidewalk and has a head-on, elastic collision with a larger 20.0-g marble sliding to the right with a velocity of magnitude 0.250 m/s.

(a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line. Take right as the positive x direction.)
m/s (smaller marble)
m/s (larger marble)

(b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each marble.
kg·m/s (smaller marble)
kg·m/s (larger marble)

(c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble.
J (smaller marble)
J (larger marble)

Explanation / Answer

let speed of 10 g 10rble =Va
Vb = speed of the 20 g
intially Vao= -.5m/s
Vbo= .25 m/s

Solving for the final velocity of a

Vaf=[(Ma-Mb)/(Ma+Mb)]Vao + [(2Mb)/(Ma+Mb)]Vbo
just imput the values
Vaf=[(10-20)/(10+20)](-0.5) + [(2*20)/(10+20)]0.25
Vaf= 0.5 m/s
0.500 m/s right

Solving for the velocity of b:

Vbf=[(2Ma)/(Ma+Mb)]Vao +[(Mb-Ma)/(Ma+Mb)]Vbo
Vbf=[(2*10)/(10+20)](-0.5) +[(20-10)/(10+20)]*0.25
= -0.25 m/s
= 0.25m/s left
b) change in momentum:
for 10 g ball , change in momemtum = 10(0.5 - (-0.5)) = 10 kg·m/s (smaller marble)

for 20 g ball , change in momemtum = 20(-0.25 - (0.25)) = -10 kg·m/s (larger marble)

c) change in kinetic energy :
for 10 g ball , change in K.E = 0.5*10(0.5^2 - (-0.5)^2) = 0 kg·m/s (smaller marble)

for 20 g ball , change in K.E = 0.5*20((-0.25)^2 - (0.25)^2) = 0 kg·m/s (larger marble)

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