A 10.0-g marble slides to the left with a velocity of magnitude 0.550 m/s on the

Question

A 10.0-g marble slides to the left with a velocity of magnitude 0.550 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 0.200 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

We know that the from the conservation of linear momentum.
m1v1i + m2v2i = m1v1f + m2v2f---------------------------1

Where m1 and m2 are the mass of the large and small marble respectively.vi and vf are the initial and final velocity respectively.
The Collision is head on AND elastic, so conservation of kinetic energy also come into picture and it simplifies as,
v2f - v1f = v1i - v2i ------------------------------------2
First let us take the 20g marble, let that one be= m1v1 since it is positive.

Solve for v2f and substitute into equation one
v2f = v1(i+f) - v2i
Substitute
m1v1i + m2v2i = m1v1f + m2(v1i + v1f - v2i)
v1f(m1+m2) = m1v1i + 2m2v2i - m2v1i
Divide LHS and RHS by m1+m2
=>v1f = [m1v1i + 2m2v2i - m2v1i]/(m1+m2)

Substituting,

v1f = [10*0.550 + 2*20*0.200 - 20*0.550]/(30)=-0.12 m/s
Direction is opposite, so it carries negative sign.
The 10g ball can be solved using conservation of momentum after you solve the above equation.
m1v1i + m2v2i = m1v1f + m2v2f
v2f = (m1v1 + m2v2 - m1v1f)/m2
v2f = (20*0.20 + 10*0.550 - 20*0.12)/10=0.66m/s
Next we need to find the change in momentom,i.e. dP, or the change in momentums of mass m2v2f - m2v2i ; use this after solving for v2f in the above.

Same thing here. m1v1f - m1v1i.
so,dP=m1v1f - m1v1i=20*0.12-20*0.20=-1.6 kg.m/s

For the 10 kg marble,dP2= m2v2f - m2v2i=1.1 kg.m/s
change in kinetic energy: We have the equation actually, so
dE=1/2 m1v1f2 - 1/2 m1v1i2

dE=0.5*20*0.122-0.5*20*0.202=-0.256J

for the second body,

dE=1/2 m2v2f2 - 1/2 m2v2i2

dE2=0.5*10*0.662-0.5*10*0.0.5502=0.665J

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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