jemarcus1999
Answered

A canoeist on a lake throws a 0.145 kg baseball at 20.0 m/s to a friend in a different canoe. Each canoe with its contents has a mass of 120 kg, and is at rest initially. How fast is the first canoe moving after the pitch, assuming no friction with the water?

Verified Answer :

MissPhiladelphia

this can be solve using momentum, where momentum = mv
where m is the mass of the object
and v is the velocity of the object

m1v1 + m2v2 = mfvf
( 0.145 kg )( 20 m/s ) + ( 120 kg )( 0 m/s) = ( 120 + 0.145 kg)vf
solve for vf

2.9 = 120.145vf
vf = 0.024 m/s

Answer

Explanation :

It is given that:

mass of baseball, \(m_1=0.145\ Kg\)

mass of canoe,  \(m_2=120\ Kg\)

velocity of baseball,  \(u_1=20\ m/s\ Kg\)

velocity of canoe, \(u_1=0\)

Let v is the velocity of the first canoe moving after the pitch.

Now, using conservation of momentum as

\(m_1u_1+m_2u_2=(m_1+m_2)v\)

\(( 0.145 kg )( 20 m/s ) + ( 120 kg )(0) = ( 120 + 0.145 kg)v\)

\(2.9\ kgm/s^2+0=120.45\ kg\ v\)

\(v=0.024\ m/s\)

Hence, it is the required solution.

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