A 30kg person is standing on a frictionless surface and someone throws a 10kg object at them with a speed of 20 meters per second. How fast will the person and the object be moving after they catch the object?

We can solve this problem by using the law of conservation of momentum. In fact, in absence of external forces (no friction), the total momentum of the system person+object must be conserved before and after the collision. Therefore we can write

[tex]p_i = p_f\\m_1 u_1 + m_2 u_2 = (m_1+m_2)v[/tex]

where:

[tex]m_1 = 30 kg[/tex] is the mass of the person

[tex]u_1 = 0 m/s[/tex] is the initial velocity of the person

[tex]m_2 = 10 kg[/tex] is the mass of the object

[tex]u_2 = 20 m/s[/tex] is the initial velocity of the object

[tex]v[/tex] is the final combined velocity of the person+object

Re-arranging the equation and substituting the values, we find the final speed:

[tex]v=\frac{m_1 u_1 + m_2 u_2}{m_1+m_2}=\frac{0+(10)(20)}{30+10}=5 m/s[/tex]

So, their final speed is 5 m/s.

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