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A ball of mass 0.220 kg that is moving with a speed of 7.5 m/s collides head-on and elastically with another ball initially at rest. Immediately after the collision, the incoming ball bounces backward with a speed of 3.8 m/s. (a) Calculate the velocity of the target ball after the collision, and (b) the mass of the target ball.

User Sdayal
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2 Answers

3 votes

Final answer:

To find the velocity of the target ball after an elastic collision and the mass of the target ball, conservation of momentum and kinetic energy must be applied. Without additional information or numerical values for the final velocity, we cannot provide an exact answer as the calculation requires specific data on the velocity after the collision.

Step-by-step explanation:

To calculate the velocity of the target ball after the collision and the mass of the target ball when a 0.220 kg ball collides elastically and the incoming ball bounces backward, we must apply the conservation of momentum and the conservation of kinetic energy which are both principles in elastic collision scenarios. We have the following data:

  • Mass of ball 1 (m1) = 0.220 kg
  • Initial velocity of ball 1 (u1) = 7.5 m/s
  • Final velocity of ball 1 (v1) = -3.8 m/s (negative because it's moving in the opposite direction after collision)
  • Mass of ball 2 (m2) = unknown
  • Initial velocity of ball 2 (u2) = 0 m/s (at rest)
  • Final velocity of ball 2 (v2) = unknown

To solve for v2, we apply the conservation of momentum:

Initial momentum = Final momentum

m1 * u1 + m2 * u2 = m1 * v1 + m2 * v2

For part (b), we need to use the conservation of kinetic energy since the collision is elastic:

Initial kinetic energy = Final kinetic energy

½ * m1 * u1^2 + ½ * m2 * u2^2 = ½ * m1 * v1^2 + ½ * m2 * v2^2

Using these equations, we can solve for v2 and m2. However, since we need a specific answer and no numerical value is provided for the velocity of the target ball, we can't calculate the exact numbers without additional data assuming a perfectly elastic collision.

User Borgy Manotoy
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7.5k points
5 votes

Answer:

Step-by-step explanation:

Given that,

Mass of ball 1

M1= 0.22kg

Initial Speed of ball 1

U1 = 7.5m/s

Initially ball 2 is at rest

U2 = 0m/s

Let mass of ball 2 be M2

Velocity of ball 1 after collision

V1 = 3.8m/s

A. Velocity of ball 2 after collision

Let the velocity of ball two after collision be V2

Using conservation of energy of momentum, then the formula reduces to

U1—U2 = —(V1 —V2)

Note that V1 is moving in the negative direction.

Check attachment

7.5—0 = —(—3.8 —V2)

7.5 = 3.8+V2

V2= 7.5—3.8

V2 = 3.7 m/s

B. Mass of ball 2= M2

Using conservation of momentum

P(before) = P(after)

M1•U1 + M2•U2 = —M1•V1 + M2•V2

0.22×7.5 + 0= —0.22 × 3.8 + 3.7•M2

1.65 = —0.836 + 3.7•M2

3.7•M2 = 1.65+0.836

M2 = 2.486/3.7

M2 = 0.67 kg

A ball of mass 0.220 kg that is moving with a speed of 7.5 m/s collides head-on and-example-1
User Samumaretiya
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7.9k points