Question

Alan hits a hockey puck of mass 0.3kg across a hockey rink with a speed of 23m/s to the south

Answer 1

a.

The momentum is given by:

`\begin{gathered} p=mv \\ where: \\ m=0.3kg \\ v=23m/s \\ so: \\ p=0.3(23) \\ p=6.9kg(m)/(s) \end{gathered}`

b.

We can use the impulse formula:

`\begin{gathered} F\cdot\Delta t=m\cdot\Delta v \\ so: \\ F=m\cdot(\Delta v)/(\Delta t) \\ \\ F=0.3\cdot(23)/(0.1) \\ F=69N \end{gathered}`

c.

`\begin{gathered} I=m\Delta v \\ where: \\ \Delta v=vf-vo=21-23=-2m/s \\ so: \\ I=0.3(-2) \\ I=-0.6kg\cdot m/s \end{gathered}`

d.

Using conservation of momentum:

`m1u1+m2v1=m1u2+m2v2`

Where:

`\begin{gathered} pi=m1u1+m2v1 \\ m1=0.3kg \\ u1=21m/s \\ m2=2kg \\ v1=0m/s \\ so: \\ pi=0.3(21)+2(0) \\ pi=6.3kg\cdot m/s \end{gathered}`

e.

Now, we can solve for the right hand side of the equation of the conservation of momentum:

`\begin{gathered} 6.3=m1u2+m2v2 \\ v2=(6.3-m1u2)/(m2) \\ where: \\ m1=0.3kg \\ m2=2kg \\ u2=4m/s \\ so: \\ v2=(6.3-(0.3)(4))/(2) \\ v2=2.55m/s \end{gathered}`