Physics, 07.02.2021 01:00 moneyyfletcher

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?

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Answer from: libertycooper

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

General Formulas and Concepts:

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets Parenthesis Exponents Multiplication Division Addition 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 momentumExplanation:

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

Substitute in variables [Law of Conservation of Momentum]: $\displaystyle (7.00 \ kg) \overline{v_{1i}} + (75.0 \ kg)(0 \ \frac{m}{s}) = (7.00 \ kg + 75.0 \ kg)(3.00 \ \frac{m}{s})$ Multiply: $\displaystyle (7.00 \ kg) \overline{v_{1i}} + 0 = 246 \ kg \cdot \frac{m}{s}$ Simplify: $\displaystyle (7.00 \ kg) \overline{v_{1i}} = 246 \ kg \cdot \frac{m}{s}$ [Division Property of Equality] Isolate unknown: $\displaystyle \overline{v_{1i}} = \frac{246}{7} \ \frac{m}{s}$ [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 from: blueheeter1461

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 kgv₂ = 0 m/s

Right side of the equation:

m₁ = 7.00 kg v₁ = 3.00 m/sm₂ = 75.0 kgv₂ = 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.

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