Question

A 2 kg basketball is thrown towards a stationary 1.34 kg kickball with a velocity of 2.15 m/s. After impact the kickball moves forward at a rate of 1.5 m/s. What is the velocity of the basketball after the collision?

Answer 1

1.15 m/s

Step-by-step explanation

Given:

• The mass of the basketball, m₁ = 2 kg

,

• The mass of the kickball, m₂ = 1.34 kg

,

• The initial velocity of the basketball, u₁ = 2.15 m/s

,

• The initial velocity of the kickball, u₂ = 0 m/s

,

• The final velocity of the kickball, v₂ = 1.5 m/s

Unknown:

• The final velocity of the basketball, v₁

By the law of conservation of momentum,

`m_1u_1+m_2u_2=m_1v_1+m_2v_2`

Solving for v₁,

`v_1=(m_1u_1+m_2u_2-m_2v_2)/(m_1)`

Replace with the known values and solve,

`v_1=(2kg\cdot2.15m/s+1.34kg\cdot0m/s-1.34kg\cdot1.5m/s)/(2kg)=(2.29kg\cdot m/s)/(2kg)\approx1.15m/s`

Hence, the velocity of the basketball after the collision is 1.15 m/s, rounded to the nearest hundredth.