Question

A canoe in a still lake is floating North at `5 m/s`. An object with a mass of `20 kg` is thrown East at `1/4 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?

Answer 1

The canoe moves off at `5.6"m"//"s North"`

Explanation:

This problem can be solved using (2-D) momentum conservation.

• The equation for momentum: `vecp=mvecv`

What we are given is an example of an explosion.

• The opposite of a collision, an explosion occurs when the particles of a system move apart from each other after a brief, intense interaction.

Momentum conservation:

`DeltavecP=0`

`=>color(darkblue)(vecP_f=vecP_i)`

For an explosion with two masses afterward, we will have:

`color(darkblue)(m_iv_i=m_1v_(f1)+m_2v_(f2))`

We are given the following information:

• `|->m_i=200"kg"`
• `|->v_i=5"m"//"s North"`
• `|->m_1=200-20=180"kg"`
• `|->m_2=20"kg"`
• `|->v_2=1/4"m"//"s East"`
• `|->theta_1=90^o`
• `|->theta_2=0^o`

Note that I am taking the positive x-axis to be my reference point for all angles.

Because the velocities given occur in different directions, we will have to decompose the vectors into their parallel (x, horizontal) and perpendicular (y, vertical) components. Therefore, we will also have components for momentum.

• Initially, the canoe and mass are combined and travel North, i.e. `90^o`. Therefore, the velocity has only a perpendicular component. This gives:

`v_i=v_y=5"m"//"s"`

Therefore, the momentum before the explosion is:

`P_i=P_(iy)=(200"kg")(5"m"//"s")`

`color(darkblue)(=1000"kgm"//"s")`

And `P_(ix)=0`.

• We know that the object is thrown East, i.e. `theta=0^o`. Therefore, the velocity has only a perpendicular component. This gives:

`v_(2f)=v_(2x)=1/4"m"//"s"`

So we have:

`color(darkblue)(P_(fx)=(180"kg")v_(1xf)+(20"kg")(1/4"m"//s"))`

`color(darkblue)(P_(fy)=(180"kg")v_(1yf))`

• Note that momentum is only conserved if each component is conserved.

Therefore:

`(180)v_(1xf)+(20)(1/4)=0`

`=>180v_(1xf)=-5`

`=>v_(1xf)=-0.02bar7"m"//"s"`

`=>color(darkblue)(~~-0.028"m"//"s")`

`180v_(1yf)=1000`

`=>v_(1yf)=5.bar5"m"//"s"`

`=>~~5.56"m"//"s"`

Therefore, the magnitude of the canoe's final velocity is:

`v_"net"=sqrt(v_x^2+v_y^2)`

`=>=sqrt((-0.028)^2+(5.56)^2)`

`=>color(darkblue)(~~5.56"m"//"s")`

Note that you can approximate the parallel component to `0`.

• If you approximate `v_x=0`, then the direction is North.

• If you do not, then you can use trigonometry to find `theta`.

`tan(theta)=v_y/v_x`

`=>theta=arctan(v_y/v_x)`

`=>theta=arctan(5.56/0.028)`

`=>=89.71^o`

`=>~~90^o`.

`:.` The canoe moves off at `5.6"m"//"s North"`.