QUESTION 1 Louisa (of mass mL=55 kg) is sitting in a canoe (of mass m c = 39 kg)

Question

QUESTION 1

Louisa (of mass mL=55 kg) is sitting in a canoe (of mass mc= 39 kg) at rest in the middle of a lake. Sadly, she has lost her oar and only has a big stone (of mass ms= 5 kg) to return to the shore (distance of 200 m). Knowing about the conservation of momentum, she throws the stone horizontally from the rear of the canoe. To reach the shore in less than 30 minutes, what minimum speed (in m/s) does she need to throw the stone? Ignoring friction.

QUESTION 2

A bullet of mass 30 g is shot upward in a 1 kg block of wood. After the collision, the bullet remains in the block. If the bullet's initial velocity when entering the wooden block is 200 m/s, what is the maximum height (in meters) reached by the block with the embedded bullet? (Assume g=-9.8 m/s2, answer to 3 SF)

Explanation / Answer

Given

Q1

mL = 55 kg , mc = 39 kg, ms = 5 kg

distance s = 200 m , time is t = 30 min = 1800 s

from conservation of momentum  

the momentum of the stone = monetum of the canone with Louisa

that is  

ms*v = (mL+mc)(s/t)

v = (mL+mc)(s/t)/ms

v = (55+39)(200/1800)/5 m/s

v = 2.088 m/s

so she need to throw the stone with speed 2.088 m/s

Q2

mb = 30 g = 0.03 kg, mB = 1 kg

initial velocity of bullet is vb = 200 m/s

from conservation of momentum of the system

momentum before and after the shot is equal

mb*vb +mB*vB = (mb+mB)V

V = (mb*vb +mB*vB ) /(mb+mB)

v = (0.03*200+1*0)/(1+0.03) m/s

v = 5.8252 m/s

now the bullet block system has kinetic energy reaches the system to a height h

where the energy will be stored in the form of gravitational potential energy

0.5*(mb+mB)v^2 = (mb+mB)*g*h

h = 0.5*v^2/g

h = 0.5*5.8252^2/9.8 m

h = 1.731 m

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