A10.0g 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.)
-find the magnitude of the velocity of 30.0 g marble after the collision.
-find the magnitude of the velocity of 10.0 g marble after the collision.
-calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision) for 30.0 g marble.
-calculate the change in momentum for 10.0 g marble.
-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.
-calculate the change in kinetic energy for 10.0 g marble

1. The final velocity of the 30.0 g marble is 0.100 m/s to the left.

2. The final velocity of the 10.0 g marble is 0.500 m/s to the right.

3. The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

4. The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

5. The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J  

6. The change in kinetic energy for the 10.0 g marble is 4.5 × 10⁻⁴ J

Explanation:

Hi there!

Since the collision is elastic both the momentum and kinetic energy of the system comprised by the two marbles is conserved, i.e., it remains constant after the collision.

momentum before the collision = momentum after the collision

mA · vA + mB · vB = mA · vA´ + mB · vB´

Where:

mA and vA = mass and velocity of the 10.0 g marble.

mB and vB = mass and velocity of the 30.0 g marble.

vA´ and vB´ = final velocities of marble A and B respectively.

The kinetic energy of the system is also conserved:

kinetic energy before the collision = kinetic energy after the collision

1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

Then, replacing with the available data:

mA · vA + mB · vB = mA · vA´ + mB · vB´

0.010 kg · (-0.400 m/s) + 0.030 kg · 0.200 m/s = 0.010 kg · vA´ + 0.030 kg · vB´

2 × 10⁻³ kg · m/s =  0.010 kg · vA´ + 0.030 kg · vB´

Solving for vA´

0.2 kg · m/s - 3 kg · vB´ = vA´

Now, using conservation of the kinetic energy:

1/2 mA · vA² + 1/2 mB · vB² = 1/2 mA · (vA´)² + 1/2 mB · (vB´)²

0.010 kg · (-0.400 m/s)² + 0.030 kg · (0.200 m/s)² = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

Replacing vA´:

2.8 × 10⁻³ kg · m/s = 0.010 kg · (vA´)² + 0.030 kg · (vB´)²

2.8 × 10⁻³ kg · m/s = 0.010 kg · (0.2 kg · m/s - 3 kg · vB´)² + 0.030 kg · (vB´)²

(I will omit units from this point for more clarity in the calculations)

2.8 × 10⁻³  = 0.010  (0.2 - 3 · vB´)² + 0.03 · (vB´)²

2.8 × 10⁻³ = 0.010(0.04 - 1.2 vB´ + 9(vB´)²) + 0.03(vB´)²

divide by 0.01 both sides of the equation:

0.28 = 0.04 - 1.2 vB´ + 9(vB´)² + 3(vB´)²

0 = -0.28 + 0.04 - 1.2 vB´ + 12(vB)²

0 = -0.24 - 1.2 vB´ + 12(vB)²

Solving the quadratic equation:

vB´= 0.200  m/s

vB´ = -0.100  m/s

The first value is discarded because it is the initial velocity. Then, the final velocity of the 30.0 g marble is 0.100 m/s to the left.

The velocity of the 10.0 g marble will be:

0.2 kg · m/s - 3 kg · vB´ = vA´

0.2 kg · m/s - 3 kg · (-0.100 m/s) = vA´

vA´ = 0.500 m/s

The final velocity of the 10.0 g marble is 0.500 m/s to the right.

The change in momentum of the 30.0 g marble is calculated as follows:

Δp = final momentum - initial momentum

Δp = 0.030 kg · (-0.100 m/s) -(0.030 kg · 0.200 m/s) = -9.00 × 10⁻³ kg · m/s

The change in momentum for the 30.0 g marble is -9.00 × 10⁻³ kg · m/s

The change in momentum of the 10.0 g marble is calculated in the same way:

Δp = final momentum - initial momentum

Δp = 0.010 kg · 0.500 m/s -(-0.010 kg · 0.400 m/s) = 9.00 × 10⁻³ kg · m/s

The change in momentum for the 10.0 g marble is 9.00 × 10⁻³ kg · m/s

The change in kinetic energy for the 30.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.030 kg · (-0.100 m/s)² - 1/2 · 0.030 kg · (0.200 m/s)²

ΔKE = -4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is -4.5 × 10⁻⁴ J

The change in kinetic energy for the 10.0 g marble will be:

ΔKE = final kinetic energy - initial kinetic energy

ΔKE = 1/2 · 0.010 kg · (0.500 m/s)² - 1/2 · 0.010 kg · (-0.400 m/s)²

ΔKE = 4.5 × 10⁻⁴ J

The change in kinetic energy for the 30.0 g marble is 4.5 × 10⁻⁴ J

an energy transformation flow diagram is shown

where is it ?

a.   mass number from atomic number