A 10.0-g marble slides to the left with a velocity of magnitude 0.450 m/s on the
ID: 1400894 • Letter: A
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.