A 10.0-g marble slides to the left at a speed of 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 at a speed of 0.200 m/s.

(a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.)
(b) Calculate the change in momentum (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble.
(c) Calculate the change in kinetic energy (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

a) VA = 0.500m/s to the right

VB = 0.100m/s to the left

b) ◇VA = 9.00gm/s

◇VB = -9.00gm/s

c) ◇KA = 0.45×10^-3J

◇KB = -0.45×10^-3J

Explanation:

aking initial directions of the proton as the positive

Vpq= 0.400m/s

Vp2 = -0.200m/s

In elastic collision,there is no external net force on the system, the total momentum before the collision is equal to the total momentum after the collision

P1 = P2

In the collision between two particles (A and B), we get:

PA1 + PB1 = PA2 + PB2

Where the momentum of a particle is given by:

P = mv

The collision equation will be:

MaVA + MBVB= MAVA2 + MBVB2 ... eq1

In elastic collision between two particles, the relative velocities before and after the collision have the same magnitude but opposite signs.VA1 - VB1 = VB2 - VA2 ...eq2

Also , VA1 -VB1 = VB2 - VA2 ...eq3

a) Using eq 3

(-0.400)- ('0.200) = VB2 - VA2

VB2 - VA2 = -0.600m/s ...eq4

Substituting into eq2

(10.0)×(-0.400) + (30.0)×(0.200)

= 10 × VA2 + 30× VB2

10VA2 + 30VB2 = 2.00

Dividing by 10

VA2 + 3 VB2 = 2.00 ...eq5

Add eq 4 and eq5 to eliminate VA2 and then solve for VB2

4VA2 = -0.400

VB2 =0.100m/s to the left

Put VB = 0.100 Iin eq 4

- 0.100 - VA2 = -0.600

VA2 = 0.500m/s to the right

b) Calculating the momentum of each particle first,gives:

PA1 = MAVA = 10 × ('0.400) = -4.00gm/s

PB1 = MBVB1 = 30 × 0.200 = 6gm/s

PA2 = MAVA2 = 10 × 0.500 = 5gm/s

PB2 = MBVB2 = 30 ×(-0.100) = -3gm/s

Therefore change in momentum of the small marble = ◇VA = VA2 - VA1 = 5.00 -(-4.00)

◇VA = 9.00gm/s

c) Calculating the kinetic energies of the particles:

KA1 = 1/2MAVA1^2 = 1/2×0.01×0.400^2 = 0.8×10^3J

KB1 = 1/2MBVB1^2 = 1/2× 0.03×0.200 = 0.3×10^-3J

KA2 = 1/2MAVA2 = 1/2 × 0.01× 0.500^2= 1.25×10^3J

KB2 = 1/2MBVB2 =1/2× 0.03×0.100^2= 0.15×10^-3J

Change in KA = KA2 - KA1

Change in KA = (1.25×10^-3)-(0.8×10^-3)

= 0.45×10^-3J

For the large particle ◇KB = KB2 - KB1

(0.15 ×10&-3)-(0.3×10^-3)

◇KB = -0.45×10^-3J

The change in KE has the same magnitude but different sign.