A 10.0-g marble slides to the left with a velocity of magnitude 0.450 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.450 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

(a)Small marble mass (M1)= 10 g and it s velocity (V1)= - 0.45 m/s (negative sign shows the direction)
Large marble mass (M2)= 20 g and its velocity (V2)= 0.25 m/s
Conservation of momentum
M1V1 + M2V2 = M1V1f+ M2V2f
10*(-0.45)+(20*.25) = 10V1f + 20V2f
10V1f + 20V2f = 0.5
Conservation of kinetic energy
(1/2)M1V12  + (1/2)M2V22 = (1/2)M1V1f2  + (1/2)M2V2f2
Replace V1f by V2f in the above equation
1.6375 = 5V1f2 + 10V2f2
1.6375 = (1/5)(0.25 - 10V2f)2 + 10V2f2
on solving V2f = 0.404 m/s and V1f = -0.758 m/s
(b) Initial momentum = M1V1 + M2V2 = 0.5
final momentum =M1V1f + M2V2f = 0.5
Therefore the Change in momentum is Zero.
(c) Similarly the change in kinetic energy is zero.
Because this is elastic collision therefore the momentum and kinetic energy is conserved.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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