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)

THE ANSWER BELOW WAS ANSWERED ON CHEGG AND INCORRECT PLEASE HELP

((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.)

Explanation / Answer

Let

Larger Marble

M1=0.02 Kg ,V1f =0.25 m/s

Smaller Marble

M2 =0.01 Kg ,V2f =-0.45 m/s

a)

By Conservation of momentum

M1V1 + M2V2 =M1V1f + M2V2f

0.02*0.25 + 0.01*(-0.45) =0.02V1f + 0.01V2f

2V1f + V2f =0.05--------------------1

Since the collision is elastic

V2f - V1f = -(V1-V2)

V1f - V2f =(-0.25-0.45)

V1f - V2f =-0.7 ---------------------2

Solving 1 and 2 we get

V1f=-0.2167 m/s(larger marble)

V2f=0.4833 m/s (Smaller marble)

b)

For Marble 1

P1 =M1(V1f-V1) =0.02(-0.2167-0.25)

P1=-0.009333 Kg-m/s (Larger Marble)

For Marble 2

P2 =M2 (V2f-V2) =0.01*[0.4833-(-0.45)]

P2=0.009333 Kg-m/s (Smaller Marble)

c)

For Marble 1

KE1 =(1/2)M1(V1f2-V12) =(1/2)*0.02*[(-0.2167)2-0.252]

KE1 =-1.5556*10-4 J (Larger Marble)

For Marble 2

KE2 =(1/2)M2(V2f2-V22) =(1/2)*0.01*(0.48332-0.452)

KE2 =1.5556*10-4 J (Smaller Marble)

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