For this problem, assume that +X is east and +y is north. A 5.00 kg block travel

Question

For this problem, assume that +X is east and +y is north. A 5.00 kg block traveling 4.00 m/s due east across a horizontal frictionless floor collides with a 3.00 kg block traveling 10.0 m/s due south. The two blocks stick together after the collision. The final velocity of the two blocks has x and y components given by The magnitude and direction of the final velocity is given by The net impulse acting on the 5.00 kg block during the collision has x and y components given by The direction of the average net force acting on the 5.00 kg block during the collision must have a direction that is

Explanation / Answer

Here ,

for final x-component ,

Vx *(5 + 3) = 5 * 4

Vx = 2.5 m/s

for final y-compenent ,

Vy *(3 + 5) = -3 *10

Vy = -3.75 m/s

Now,

magnitude of velocity = sqrt(2.5^2 + 3.75^2)

magnitude of velocity = 4.51 m/s

angle = arctan(3.75/2.5) degree south of east

angle = 56.3 degree south of east

Now , for block A

Fx delt = 5*(2.5 - 4)

Fx delt = -7.5 N.s

Fy del t = 5*(-3.75 - 0)

Fy*del t = -18.75 N.s

direction of average force = 180 + arctan(18.75 / 7.5)

direction of average force = 248.2 degree north of east

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