A 10.0-g marble slides to the left with a velocity of magnitude 0.350 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.350 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

m1 = 20 g = 0.02 kg

m2 = 10 g = 0.01 kg

u1 = 0.25 m/s

u2 = - 0.35 m/s

Perfectly elastic collision

Applying energy and momentum conservation

m1 u1 + m2 u2 = m1 v1 + m2 v2

0.5 m1u1^2 + 0.5 m2 u2^2 = 0.5 m1 v1^2 + 0.5 m2 v2^2

Solving above equations we get

v1 = - 0.33 m/s

v2 = 1.516 m/s

(a) velocity of smaller marble = 1.516 m/s

velocity of larger marble = - 0.33 m/s

(b) Change in momentum of smaller marble = m2 v2 - m2 u2 = 0.01*(1.516 + 0.35) = 0.0186 kg m/s

Change in momentum of larger marble = m1 v1 - m1 u1 = 0.02*(-0.33 - 0.25) = -0.0116 kg m/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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