Question

tion 14A 37 kg child holds a 5.9 kg package in a 67 kg boat at rest, and then the child throws the packagehorizontally from the boat at a velocity of +12.4 m/s. What is the velocity of the boat after this?Vunits

Answer 1

-0.704 m/s

Step-by-step explanation:

The conservation of momentum demands that since the child + boat + box system is isolated, the initial momentum must equal the final momentum.

Now since initially, everything is at rest, the initial momentum of the system is zero.

The final momentum of the system is

`(m_c+m_b)v_1+(m_{\text{box}})v_2`

where

m_c = mass of the child

m_b = mass of the boat

m_box mass of the boat

v1 = belcity of child + boat

v2 = veloctiy of the box.

Equating the initial momentum to the final momentum gives

`(m_c+m_b)v_1+(m_{\text{box}})v_2=0`

Now in our case

m_c = 37 kg

m_b = 67 kg

v1 = unknown

m_box = 5.9 kg

v2 = 12.4

Therefore, the above equation gives

`(37+67)v_1+(5.9)(12.4)=0`

solving for v1 gives

`v_1=(5.9\cdot12.4)/(37+67)`

which evaluates to give (rounded to the nearest hundredth)

`\boxed{v_1=-0.70m/s}`

which is our answer!