Chapter 11: Q 79 (page 291)

A 1000 kg cart is rolling to the right at 5.0 m/s. A 70 kg man is standing on the right end of the cart. What is the speed of the cart if the man suddenly starts running to the left with a speed of 10 m/s relative to the cart?

Short Answer

Expert verified

Speed of the cart is 5.65 m/s

Step by step solution

Step 1. Given information is :Mass of the cart = 1000 kgSpeed of the cart = 5.0 m/s towards rightMass of the man = 70 kgSpeed of the man is 10 m/s relative to the cart towards Left

We need to find the speed of the cart if the man suddenly starts running to the left with a speed of 10 m/s relative to the cart.

Step 2. Using Law of conservation of momentum for finding the speed of the cart

Speed of the man relative to the cart = -10 m/s (negative as moving towards left)

$v_{m c} = - 10 m / s v_{m c} = v_{m} - v_{c} - 10 m / s = v_{m} - v_{c} v_{m} = - 10 m / s + v_{c}$

Using Law of conservation of momentum,

$\vec{P_{i}} = \vec{P_{f}} (m_{c} + m_{m}) v_{i} = m_{c} v_{c} + m_{m} v_{m} (m_{c} + m_{m}) v_{i} = m_{c} v_{c} + m_{m} (- 10 m / s + v_{c}) (m_{c} + m_{m}) v_{i} = - 10 m / s \times m_{m} + m_{c} v_{c} + m_{m} v_{c} (m_{c} + m_{m}) v_{i} + 10 m / s \times m_{m} = (m_{c} + m_{m}) v_{c} v_{c} = v_{i} + \frac{10 m / s \times m_{m}}{(m_{c} + m_{m})} P u t t i n g v a l u e s, v_{c} = 5 m / s + \frac{(10 m / s) (70 k g)}{1000 k g + 70 k g} v_{c} = 5 m / s + \frac{700 k g m / s}{1070 k g} v_{c} = 5.65 m / s$

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A 1000 kg cart is rolling to the right at 5.0 m/s. A 70 kg man is standing on the right end of the cart. What is the speed of the cart if the man suddenly starts running to the left with a speed of 10 m/s relative to the cart?

Short Answer

Step by step solution

Step 1. Given information is :Mass of the cart = 1000 kgSpeed of the cart = 5.0 m/s towards rightMass of the man = 70 kgSpeed of the man is 10 m/s relative to the cart towards Left

Step 2. Using Law of conservation of momentum for finding the speed of the cart

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