Question

A dry cleaner tosses a bag of laundry into a stationary 9.0kg cart with a horizontal velocity of 4.2m/s to the right. If the cart then begins moving at 3.0m/s to the right, how much does the bag of laundry weigh?

Answer 1

We can find the weight of the bag of laundry from the law of conservation of momentum:

$$m_1v_{1i} + m_2v_{2i} = (m_1 + m_2)v_f,$$

here,

$$m_1$$

,

$$m_2$$

are the masses of the bag and the cart, respectively;

$$v_{1i}$$

,

$$v_{2i}$$

are the initial velocities of the bag and the cart, respectively;

$$v_f$$

is the final velocity of the system of cart and bag.

Then, from this equation we can find the mass of the bag,

$$m_2$$

:

$$m_1(v_{1i} - v_f)=m_2(v_f - v_{2i}),$$ $$m_1 = \frac{m_2(v_f - v_{2i})}{(v_{1i} - v_f)}.$$

Let's substitute the numbers:

$$m_1 = \frac{9 kg \cdot (3 \frac{m}{s} - 0 \frac{m}{s})}{(4.2\frac{m}{s} - 3\frac{m}{s})} = 22.5 kg.$$

Finally, we can find the weight of the bag of laundry:

$$W_{bag} = m_2g = 22.5 kg \cdot 9.8 \frac{m}{s^2} = 220 N.$$

Answer:

$$W_{bag} = 220 N$$

.