Question

A 7.00 kg ball hits a 75.0 kg man standing at rest on ice. The man catches the ball. How fast does the ball need to be moving in order to send the man off at a speed of 3.00 m/s?

Answer 1

Answer:

$\displaystyle \overline{v_{1i}} = 35.1429 \ \frac{m}{s}$

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right  

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Physics

Momentum

Momentum Formula: $\overline{P} = m \overline{v}$

• P is momentum (in kg · m/s)
• m is mass (in kg)
• v is velocity (in m/s)

Law of Conservation of Momentum: $\sum \overline{P_i} = \sum \overline{P_f}$

• States that the sum of initial momentum must equal the sum of final momentum

Explanation:

Step 1: Define

[LCM] $\sum \overline{P_i} = \sum \overline{P_f}$  →  $m_{1} \overline{v_{1i}} + m_{2} \overline{v_{2i}} = (m_{1} + m_{2}) \overline{v_{f}}$

m₁ (ball) = 7.00 kg

m₂ (man) = 75.0 kg

$\overline{v_{1i}} = ?$

$\overline{v_{2i}} = 0 \ \frac{m}{s}$ (man starts from rest)

$\overline{v_{f}} = 3.00 \ \frac{m}{s}$ (the ball and the man are one mass because the man catches and keeps the ball)

We know no energy is lost because it is a frictionless surface. The collision should be perfectly elastic.

Step 2: Solve

1. Substitute in variables [Law of Conservation of Momentum]:                    $\displaystyle (7.00 \ kg) \overline{v_{1i}} + (75.0 \ kg)(0 \ \frac{m}{s}) = (7.00 \ kg + 75.0 \ kg)(3.00 \ \frac{m}{s})$
2. Multiply:                                                                                                           $\displaystyle (7.00 \ kg) \overline{v_{1i}} + 0 = 246 \ kg \cdot \frac{m}{s}$
3. Simplify:                                                                                                          $\displaystyle (7.00 \ kg) \overline{v_{1i}} = 246 \ kg \cdot \frac{m}{s}$
4. [Division Property of Equality] Isolate unknown:                                           $\displaystyle \overline{v_{1i}} = \frac{246}{7} \ \frac{m}{s}$
5. [Evaluate] Divide:                                                                                           $\displaystyle \overline{v_{1i}} = 35.1429 \ \frac{m}{s}$

The initial speed of the ball should be approximately 35.14 m/s.

Answer 2

Answer:

35.14 m/s

Explanation:

The Law of Conservation of Momentum states that the momentum before and after a collision is the same.

• m₁ v₁ + m₂ v₂ = m₁ v₁ + m₂ v₂

Let's set the ball to have the subscript of 1 and the man to have the subscript of 2.

The initial and final mass of the ball is the same, so m₁ = 7.00 kg on both sides of the equation.

The initial velocity of the ball, v₁ on the left side of the equation, is the unknown variable we are trying to find.

The initial and final mass of the man is the same, so m₂ = 75.0 kg on both sides of the equation.

The man starts at rest, meaning that his initial velocity is v₂ = 0 m/s on the left side of the equation.

The final velocity of both the ball and the man is 3.00 m/s, so we can set v₁ and v₂ on the right side of the equation to equal 3.00 m/s.

Left side of the equation:

• m₁ = 7.00 kg
• v₁ = ?
• m₂ = 75.0  kg
• v₂ = 0 m/s

Right side of the equation:

• m₁ = 7.00 kg
• v₁ = 3.00 m/s
• m₂ = 75.0  kg
• v₂ = 3.00 m/s  

Substitute these values into the Law of Conversation of Momentum formula.

• (7.00) v₁ + (75.0)(0) = (7.00)(3.00) + (75.0)(3.00)

Multiply and simplify.

• 7.00 v₁ = 21 + 225
• 7 v₁ = 246

Divide both sides of the equation by 7.

• v₁ = 35.14 m/s

The ball needs to be moving at a speed of 35.14 m/s in order to send the man off at a speed of 3.00 m/s.