Collision Problem: Masses, Speed, Force

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SUMMARY

The discussion focuses on a physics problem involving a box of mass 6.30 kg dropped from a height of 2.10 m onto a skateboard of mass 8.70 kg, which is initially moving at 7.00 m/s. The final speed of the skateboard and box system after the collision is calculated to be 4.06 m/s. The average force exerted by the skateboard on the box during the collision, lasting 8.00 ms, requires calculating the impulse in both x and y directions, considering the change in momentum vertically and horizontally.

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Homework Statement



A box of mass 6.30 kg is dropped from a height of 2.10 m onto a super-sized skateboard of mass 8.70 kg that rolls on a horizontal surface where no friction acts. Initially the skateboard moves at 7.00 m/s.
(a) Assume that the box stays on the skateboard. What is the final speed?
(b) The collision of the box and the skateboard lasts a time of 8.00 ms. What is the average force exerted by the skateboard on the box during the time of the collision?

Homework Equations



J= Favg * t

The Attempt at a Solution


I already have found part a to equal 4.06 m/s. I just need help figuring out what to do with part b, which is to be reported in i-hat j-hat notation.
Thanks
 
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I'll admit that I am not completely sure of this myself, but I'm thinking that you have to find the impulse in both the x and y directions. Because there will be a change in momentum vertically as the box hits the board, but also because the box ends up moving at the same speed as the board then the board must also be applying a force (or impulse) to the box in the horizontal direction.
Then recombine the components to find the total impulse vector and solve for t.
 

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