A mass (MA=3kg) slides down a frictionless ramp of height H=12m and collides wit

Question

A mass (MA=3kg) slides down a frictionless ramp of height H=12m and collides with a second mass (MB=5kg), moving towards mass A at vB=8m/s . Assume the collision occurs with both masses moving horizontally. ????a) (5 pts) What is the velocity of Mass A at the bottom of the ramp? vA =____________ b) (10pts) Assuming the collision is elastic, what are the velocities of the two masses after the collision? v'A =____________ v'B =____________ c) (10pts) How far back up the ramp does Mass A travel? Hnew=____________

Explanation / Answer

(a) velocity of mass A: conservation of energy: mgh = (1/2)mv ^2 --> v= sqrt(2gh) = sqrt(2*9.8*12)=15.33 m/s (b) elastic collision: x is speed of A y is speed of B conservation of linear momentum: (3)(15.33) + 5(-8) = 3x +5y conservation of energy: (3)(15.33)^2 + 5(-8)^2 = 3x^2 +5y^2 solving the equations: v_a = -13.82 m/s v_b= 9.497 m/s if this dosen't work alternate the sign, because i don't know what sign convention you are using. height up the ramp: h= (1/2)v^2/g = (1/2)*(13.82^2)/9.81 = 9.734 m

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