Pad 10:08 AM O sakai, rutgers. edu : ELE Bonus Points Exercise- R WebReg i View

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Pad 10:08 AM O sakai, rutgers. edu : ELE Bonus Points Exercise- R WebReg i View / Print S × G what is your t crit value C www.webassign.net/web/Student/Assignment-Responses/last?dep-17486714 0 (a) the magnitude of the impulse delivered to the ball (b) the average force exerted on the ball kN Need Help? Read ItWatch It 3. + -/2 points SerPSET9 9.P.041. MI My Notes Ask Your Teacher A billiard ball moving at 6.00 m/s strikes a stationary ball of the same mass. After the collision, the first ball moves at 4.97 m/s at an angle of 34.0 with respect to the original line of motion. Assuming an elastic collision (and ignoring friction and rotational motion), find the struck ball's velocity after the collision magnitude directionn m/s o (with respect to the original line of motion) Need Help? Read ItMaster It 4. -/1 points SerPSET9 10.P.041 My Notes Ask Your Teacher The figure below shows a side view of a car tire before it is mounted on a wheel. Model it as having two sidewalls of uniform thickness 0.630 cm and a tread wall of uniform thickness 2.50 cm and width 18.2 cm. Assume the rubber has a uniform density 110 x 103 kalm3 Find its moment of inertia ahout an axis nernendicular to the nage through its center

Explanation / Answer

given:

moving ball:

mass = m, vi = 6 m/s

stationary ball:

mass = m, vi = 0 m/s

conservation of momentum:

x-direction: (m)(6) + (m)(0) = m(4.97 cos34) + m(vfx)

6m = 4.12m + m(vfx)

vfx = 1.88 m/s

y-direction:

m(0) + m(0) = m(4.97 sin34) + m(vfy)

-2.78 m = mvfy

vfy = -2.78 m/s

magnitude:

vf = sqrt[vfx^2 + vfy^2]

vf = 3.36 m/s

direction:

arctan(vfy/vfx) = -55.93 degrees

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