A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the
ID: 1478583 • Letter: A
Question
A 10.0-g marble slides to the left with a velocity of magnitude 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head-on, elastic collision with a larger 30.0-g marble sliding to the right with a velocity of magnitude 0.200 m/s. Let +x be to the right. (Since the collision is head-on, all the motion is along a line.) (Figure 1)
A-Calculate the change in momentum(that is, the momentum after the collision minus the momentum before the collision) for 30.0-g marble.
B-Calculate the change in momentum for 10.0-
g marble.
C-Calculate the change in kinetic energy(that is, the kinetic energy after the collision minus the kinetic energy before the collision) for 30.0-
g marble.
D-Calculate the change in kinetic energyfor 10.0-
g marble
Explanation / Answer
the final velocity V1=(m1-m2)/(m1+M2) *u1 + 2m2/(m1+m2)*u2
m1=30 g=0.03kg; m2=10 g=0.01Kg
u1=0.200 m/s ; u2=-0.400 m/s***********************on substuting data****************
v1=final velocity of first body; v2=final velocity of second body
V1=0.1-0.2=-0.1 m/s
v2=2m1u1/(m1+m2) + (m2-m1) u2 /(m1+m2)
v2=0.3-(-0.2)
v2=0.5 m/s
****************************
a)change in momentum of first body=m1(v1-u1)
=0.03(-0.1-0.2)
=-0.009 kgm/s
b)change in momentum of second body=m2(v2-u2)
=0.01*(0.5-(-0.4))
=0.009 kgm/s
c)0.5*m1(v1^2-u1^2) =0.5*0.03*(-0.01)=-0.00015 J
d)0.5*m2*(v2^2-u^2)=0.5*0.01(0.24)=0.0012 J
****************************
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