MHB Momentum Change in Hockey Ball: Mass 0.2kg, Speed 8m/s to 5m/s

[Shah 72]

A hockey ball of mass 0.2kg is hit so that its initial speed is 8 m/s. The ball travels in a horizontal straight line with acceleration given by a= - 0.5- kt where t is the time in seconds measured from when the ball was hit. After 2s the ball has traveled 41/3 m. It is then intercepted by a players from the other team. This player hits the ball so that its direction of travel is reversed and its speed is now 5 m/s. Show that when the ball is hit by the second player it's momentum changes in magnitude by 2 Ns

 

[DaalChawal]

Integrate the above expression to get velocity at t=2 sec and then apply $m(-5 ) - m(v_{t=2})$ assuming ball is traveling in positive x- direction initially.

 

[Shah 72]

Integrate the above expression to get velocity at t=2 sec and then apply $m(-5 ) - m(v_{t=2})$ assuming ball is traveling in positive x- direction initially.

So after integrate I get v= -0.5t- kt^2/2 +c
When t=0, v= 8m/s
So v= -0.5t -kt^2/2+8

 

[Shah 72]

Integrate the above expression to get velocity at t=2 sec and then apply $m(-5 ) - m(v_{t=2})$ assuming ball is traveling in positive x- direction initially.

I don't understand how to calculate after this

 

[skeeter]

$v(t) = -0.5t - \dfrac{kt^2}{2} + 8$

$\displaystyle \int_0^2 v(t) \, dt = \dfrac{41}{3}$

evaluate the integral, solve for $k$, then determine the value of $v(2)$

$\Delta \vec{p} = m[-5 - v(2)]$

 

[Shah 72]

$v(t) = -0.5t - \dfrac{kt^2}{2} + 8$

$\displaystyle \int_0^2 v(t) \, dt = \dfrac{41}{3}$

evaluate the integral, solve for $k$, then determine the value of $v(2)$

$\Delta \vec{p} = m[-5 - v(2)]$

Thank you very much!