jemarcus1999
contestada

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?

Respuesta :

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, [tex]m_1=0.145\ Kg[/tex]

mass of canoe,  [tex]m_2=120\ Kg[/tex]

velocity of baseball,  [tex]u_1=20\ m/s\ Kg[/tex]

velocity of canoe, [tex]u_1=0[/tex]

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

Now, using conservation of momentum as

[tex]m_1u_1+m_2u_2=(m_1+m_2)v[/tex]

[tex]( 0.145 kg )( 20 m/s ) + ( 120 kg )(0) = ( 120 + 0.145 kg)v[/tex]

[tex]2.9\ kgm/s^2+0=120.45\ kg\ v[/tex]

[tex]v=0.024\ m/s[/tex]

Hence, it is the required solution.

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