Two balls collide in a head?on, completely elastic collision. Ball #1 is 0.50 kg

Question

Two balls collide in a head?on, completely elastic collision. Ball #1 is 0.50 kg and is travelling with a speed of 4.3 m/s in the positive x?direction. Ball #2 is 0.50 kg and is travelling with a speed of 1.8 m/s in the negative x?direction. A. What will the speed and direction of each ball be after the collision? B. What is the Kinetic Energy (KE) of each ball, and the total KE before the collision? What is the KE of each ball, and the total KE after the collision? Explain. C. After the collision, Ball #2 bounces off a nearby wall and rebounds with

Explanation / Answer

m1v1+m2v2 = m1v1' + m2v2' m1v1' +m2v2' = 2.5 m1=m2 =.5kg v1'+v2' = 2.5 ---eqn(1) intial K.E = final K.E 1/2mv1^2 + 1/2mv2^2 = 1/2mv1'^2 + 1/2mV2'^2 v1^2 + v2^2 = v1'^2 + v2'^2 v1'^2 + v2'^2= 21.73 from eqn1 (2.5-v1')^2 + v1'^2 = 21.73 body 1 after collision v1' = -1.8 =>-x direction body 2 after collision v2' = 4.3 => +x direction k.E of each ball = ball1 = .81 J ball 2 = 4.622J t.E = 5.43J V2' = 4.3m/s and bounce back by = 2.15 in 44*10^-3sec v= u+at ==> a = -146.59m/s^2 ===> avg force = ma = -146.59*.5 = -73.29N

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