A canoe in a still lake is floating North at #8 m/s#. An object with a mass of #20 kg# is thrown North East at #3 m/s#. If the mass of the canoe was #200 kg # before the object was thrown, what is the new speed and direction of the canoe?

2 Answers
Mar 14, 2018

See principle below I leave the final math to you.

Explanation:

Work this out as a momentum problem.
Momentum = #m*v#
Momentum of canoe = #200*8 = 1600# kg.m/s directed north
Momentum of rock = #20*83= 60# kg.m/s directed NE ( 45 deg )

Resolve momentum vector of rock into north and east component:

North component: #60 sin45 =60 *0.707 = 42.42#
East component :#60 sin45 =60 *0.707 = 42.42#

So combined canoe + rock momentum is two vectors:

North: #1642.42 #
East : #42.42#

Resultant vector magnitude is: #sqrt(1642.42^2 + 42.42^2)#
Direction: #sin^-1(42.42/1642.42)#

Mar 16, 2018

The new speed is 7.79 m/s and the direction is #1.56^@# West of North.

Explanation:

Momentum is conserved.
Momentum before = momentum after

The previous equation and the following equation is the statement of conservation of momentum in this case.

#(m_1+m_2)*u = m_1*v_1 + m_2*v_2#
where #m_1 and m_2# are the masses of the canoe and the 20 kg object respectively, u and the 2 v's (#v_1 and v_2#) are the velocities before and after respectively. Since the canoe's course was North and the object was thrown North East, I will break the velocity of the object into North and East components.

#v_"2n" = 3 m/s*sin45 = 2.12 m/s#

The statement of the problem allow 2 different interpretations. When it says that the object was thrown North East at 3 m/s, was the 3 m/s with respect to the canoe or to the lake? If with respect to the lake is what was meant, since the canoe's velocity was 8 m/s, the object was actually thrown South East #-# with respect to the canoe. I will choose to interpret the 3 m/s NE as with respect to the canoe. Therefore I will add the canoe's 8 m/s to the above 2.12 m/s.

#v_"2e" = 3 m/s*cos45 = 2.12 m/s#

The momentum equation needs to be done separately for the North direction and the East direction.

North:
Entering the data, the momentum equation becomes

#(200 kg*+20 kg)*8 m/s = 200 kg*v_"1n" + 20 kg*(8 m/s+2.12 m/s#

Solving for #v_"1n"#

#1760 kg*m/s = 200 kg*v_"1n" + 202.4 kg*m/s#

#200 kg*v_"1n" = 1760 kg*m/s - 202.4 kg*m/s = 1558 kg*m/s#

#v_"1n" = (1558 kg*m/s)/(200 kg) = 7.79 m/s#

East:
Entering the data, the momentum equation becomes

#(200 kg*+20 kg)*0 = 200 kg*v_"1e" + 20 kg*2.12 m/s#

Solving for

#0 = 200 kg*v_"1e" + 20 kg*2.12 m/s#

#200 kg*v_"1e" = - 42.4 kg*m/s#

#v_"1e" = - (42.4 kg*m/s)/(200 kg) = -0.212 m/s#

So this component of the final velocity is actually west.

The new speed is the vector sum of #v_"1n" and v_"1e"#.

#"New speed" = sqrt(7.79^2 + (-0.212)^2) = 7.79 m/s#

#"Direction" = tan^-1(0.212/7.79) = 1.56^@# West of North.

I hope this helps,
Steve