Answer:
[tex]F = \frac{10mv_0}{11\Delta t}[/tex]
Explanation:
The change in the momentum of an object is equal to the average force applied to that object in a given interval of time. We will apply this rule to the bigger mass, who is at rest initially.
[tex]\vec{F}\Delta t = \vec{p}_2 - \vec{p}_1\\F\Delta t = 10mv_2 - 0\\F = \frac{10mv_2}{\Delta t}[/tex]
[tex]v_2[/tex] can be found by the conservation of momentum:
[tex]\vec{p}_{initial} = \vec{p}_{final}\\mv_0 + 0 = (10m + m)v_2\\mv_0 = 11mv_2\\v_2 = \frac{v_0}{11}[/tex]
Now the average force can be written in terms of v0, m, and Δt.
[tex]F = \frac{10m}{\Delta t}v_2 = \frac{10m}{\Delta t}\frac{v_0}{11}[/tex]
