Phsyics - Answer both questions while showing all of your work and I\'ll be sure
ID: 2032402 • Letter: P
Question
Phsyics - Answer both questions while showing all of your work and I'll be sure to give you an upvote. Thanks
1. A 220g ball is dropped from a height of 2.0 m, bounces on a hard floor, F and rebounds to a height of 1.5 m. The figure shows the impulse received from the floor. What maximum force does the floor exert on the ball? A 2100 kg truck is traveling east through an intersection at 2.0 m/s when it is hit simultaneously from the side and the rear. (Some people have all the luck!) One car is a 1200 kg compact traveling north at 5.0 m/s. The other is a 1500 kg midsize traveling east at 11 m/s. The three vehicles become entangled and slide as one body. What are their speed and direction just after the collision? 2.Explanation / Answer
1)given
m = 220 g = 0.22 kg
h1 =2 m
h2 = 1.5 m
speed of the ball just before striking the ground,
v1 = sqrt(2*g*h1)
= sqrt(2*9.8*2)
= 6.26 m/s
velocity, v1 = -6.26 m/s (negative represents downward direction)
velocity of ball just after bouncing, v2 = sqrt(2*g*h2)
= sqrt(2*9.8*1.5)
= 5.42 m/s
now use, Impulse = change in momentum
(1/2)F_max*delta_t = m*(v2 - v1)
F_max = 2*m*(v2 - v1)/delta_t
= 2*0.22*(5.42 - (-6.26))/(5*10^-3)
= 1028 N <<<<<<-------------------Answer
2)
let East be +x axis
let vx and vy are the componets of final velocity.
Apply conservation of momentum in x-direction
vx*(2100 + 1200 + 1500) = 2100*2 + 1200*0 + 1500*11
vx = (2100*2 + 1500*11)/(2100 + 1200 + 1500)
= 4.31 m/s
Apply conservation of momentum in y-direction
vy*(2100 + 1200 + 1500) = 2100*0 + 1200*5 + 1500*0
vy = (1200*5)/(2100 + 1200 + 1500)
= 1.25 m/s
final speed, v = sqrt(vx^2 +vy^2)
= sqrt(4.31^2 + 1.25^2)
= 4.5 m/s <<<<<<<-------------------Answer