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ID: 1287767 • Letter: C
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An object has a kinetic energy of 242 J and a momentum of magnitude 27.4 kg-m/s. Find the speed and the mass of the object. speed m/s mass kg At one instant, a 17.0-kg sled is moving over a horizontal surface of snow at 3.60 m/s. After 8.70 s has elapsed, the sled stops. Use a momentum approach to find the magnitude of the average friction force acting on the sled while it was moving. N A 3.02 kg particle has a velocity of (2.96 - 4.07 ) m/s. (a) Find its x and y components of momentum. Px = kg-m/s Py = kg-m/s (b) Find the magnitude and direction of its momentum. kg-m/s degree (counterclockwise from the axis) A baseball approaches home plate at a speed of 46.0 m/s, moving horizontally just before being hit by a bat. The batter hits a pop-up such that after hitting the bat, the baseball is moving at 60.0 m/s straight up. The ball has a mass of 145 g and is in contact with the bat for 2.20 ms. What is the average vector force the ball exerts on the bat during their interaction?Explanation / Answer
1. k.e. = mv^2 /2 = 242
P = mv =27.4
k.e / p = v/2 = 242/27.4
v = 17.66 m/s
m = 27.4 / 17.66 = 1.55 kg
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2. Impulse = change in momentum = F x time
17 ( 3.60 - 0 ) = F x 8.70
F = 7.03 N
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3. P = mv = 3.02(2.96i - 4.07j) = 8.94i - 12.29 j
Px = 8.94 kg.m/s
Py = -12.29 kg.m/s
b)
magnitude = sqrt(px^2 + py^2) = 15.20 kg.m/s
direction = 360 - tan-1(12.29/8.94) =306.03 degrees
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4. impulse = chaneg in momentum = F x time
0.145 ( 60j - 46i ) = F x 2.20 x 10-3
F = - 3031.82i + 3954j
this is the force acted on ball.
from newtons 3rd law . same magnitude but opposite direction force being exerted on bat by ball.
F = 3031.82i - 3954j N
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