Question

James m = 85.0 kg and ramon m = 59.0 kg are 20.0 m apart on a frozen pond. midway between them is a mug of their favorite beverage. they pull on the ends of a light rope stretched between them. ramon pulls on the rope to give himself a speed of 1.20 m/s. what is james's speed?

Answer 1

You can use conservation of momentum, Ramon has a momentum of m * v = 59.0 * 1.20 = 70.8. James must have that same momentum so m * v = p -->
p / m = v --> 70.8 / 85.0 = 0.83 m/s for James

Answer 2

The speed of James as the result of the pull is 0.83 m/s.

The given parameters;

• mass of James, m₁ = 85 kg
• mass of Ramon, m₂ = 59 kg
• speed of Raman, u₂ = 1.2 m/s

Apply the principle of conservation of linear momentum, since James and Ramon pull the rope at both end, momentum must be conserved.

m₁u₁ = m₂u₂

where;

• u₁ is the speed of James

$u_1 = \frac{m_2u_2}{m_1}\\\\u_1 = \frac{59 \times 1.2}{85}\\\\u_1 = 0.83 \ m/s$

Thus, the speed of James as the result of the pull is 0.83 m/s.

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