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 25.0-g marble sliding to the right with a velocity of magnitude 0.150 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 xdirection.)

(b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for each 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.

Explanation / Answer

here,

mass of small mable , ms = 0.01 kg

mass of big large marble , ml = 0.025 kg

initial velociity of small marble , us = - 0.5 m/s

initial velocity of large marble , ul = 0.15 m/s

let their final speed be vs and vl

(a)

using conservation of momentum

ms*us + ml*ul = ms*vs + ml*vl

- 0.01*0.5 + 0.025 * 0.15 = 0.01*vs + 0.025*vl ...(1)

using conservation of kinetic energy

0.5*ms*us^2 + 0.5*ml*ul^2 = 0.5*ms*vs^2 + 0.5*ml*vl^2

0.01*0.5^2 + 0.025 * 0.15^2 = 0.01*vs^2 + 0.025*vl^2 ...(2)

from equation (1) and (2)

vs = 0.43 m/s , vl = - 0.22 m/s

the final velocity of the small ball is 0.43 m/s and the large ball is -0.22 m/s

(b)

the change in the momentum for small marble , ps = ms( vs - us)

Ps = 0.01 * ( 0.43 + 0.5)

Ps = 9.3 * 10^-3 kg.m/s

the change in momentum for the large marble , pl = ml*( vl - ul)

pl = 0.025*( - 0.22 - 0.15)

pl = - 9.3 * 10^-3 m/s

(c)

the change in kinetic energy for the small mable , KEs = 0.5 * ms*( vs^2 - us^2)

KEs = 0.5 * 0.01 * ( 0.43^2 - 0.5^2)

KEs = 3.2 * 10^-4 J

the change in kinetic energy for the large mable , KEl = 0.5 * ml*( vl^2 - ul^2)

KEl = 0.5 * 0.025 * ( 0.22^2 - 0.15^2)

KEs = 3.2 * 10^-4 J

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