Three pool balls of equal mass are resting on a horizontalfrictionless surface.
ID: 1756219 • Letter: T
Question
Three pool balls of equal mass are resting on a horizontalfrictionless surface. The cue ball (Ball 1) of mass 700 g is rolledinto a stationary pool ball (Ball 2) and rebounds at 1.00 m/s inthe direction from which it came. Ball 2 does not move, but it istouching another pool ball (Ball 3), which rolls away at 4.00 m/s.Which of the following is the initial speed of the cue ball,v1i? Also, given that the encounterbetween the cue ball and cue stick lasts for only 0.0500 s, whichof the following is the average force delivered F to thecue ball by the stick?Explanation / Answer
Let Mass of the cue ball (m1) = 700 g = 0.7 kg Mass of the second ball (m2) = 0.7 kg Mass of the third ball (m3) = 0.7 kg Initial velocity of cue ball = u1 Initial velocity of the second ball (u2) = 0 Initialvelocity of third ball (u3) = 0 Final velocity of the cue ball (v1) = -1.00 m /s (as it rebounds back) Finalvelocity of the other two balls (v) = 4.00 m /s From the conservation of momentum we have : m1 u1 + m2 u2 + m3 u3 = m1v1 + (m2 + m3 ) v m1 u1 + 0 + 0 = m1 ( -1.00 ) + ( m2 + m3 ) *4.00 m /s or 0.7 kg * u1 = - 0.7 kg + ( 0.7 +0.7 ) * 4 = 4.9 or u1 = 7 m/s Hence Initial velocity of the cue ball is : u1 = 7 m/s (b) Average force is : F = m ( u2 - u1 ) /t = ( 0.7 * ( -1.00 - 7.0 ) ) / 0.050 s = -112 N Hope this helps u!Related Questions
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