A 2.50kg block slides from rest atop a 4.00m long, smooth incline which is eleva

Question

A 2.50kg block slides from rest atop a 4.00m long, smooth incline which is elevated 37 degrees above the horizontal. At the bottom of the incline the 2.50kg mass makes an elastic collision with a larger block that was sitting at rest. A). Determine the speed of the2.5kg blocck just before the collision. B) If the 2.5kg block slides 1.23m back up the incline, calculate the recoil speed of the smaller mass. C). Determine the velocity of the larger mass immediately after the collision. D) calculate the mass of the larger block. ( H=4sin37=2.4m)

Explanation / Answer

A) Speed of the 2,5 kg block just before the collision is v = sqrt(2*g*l*sin(37)) = sqrt(2*9.81*4*sin(37)) = 6.87 m/s

B) From lawof conservatio of energy

0.5*m*v^2 = m*g*l*sin(37)

recoil speed is v = sqrt(g*l*sin(37) / (0.5)) = sqrt(9.81*1.23*sin(37)/0.5) = 3.81 m/s

C)

velocity of larger mass u2 = (m1*u1)/m2 = (2.5*3.81)/(

in elastic collision
v2-v1 = u1-u2
given that v1 = 0 m/s
u1 = 6.87m/s
u2 = 3.81
then v2 =6.87-3.81 = 3.06 m/s

D) From law of conservation momentum

m1*u1 =-m2*v2

m2 = (m1*u1)/u2 = (2.5*6.87)/3.06 = 5.61 Kg

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