Two blocks of mass 2 kg and 3 kg are travelling towards each other at speeds of

Question

Two blocks of mass 2 kg and 3 kg are travelling towards each other at speeds of 2 m/s and 3 m/s, respectively. They stick after they collide. 1.Determine the speed after the collision and its direction. 2. Calculate the Kinetic energy lost in the collision above. 3. Suppose the collision in problem 1 is elastic. Determine the final speeds and directions. 4. Suppose the collision in problem 1 has coefficient of restitution equal to 0.5. Determine the final speeds and directions. 5. In the previous problem, what is Q?

Explanation / Answer

a)

m1+m2 v = m1 v1 + m2 v2

==> v = (m1 v1 + m2 v2)/(m1+m2)

v = (3*3-2*2)/(2+3) = 1 m/s

b)

Ki = 0.5 (m1 v1^2 + m2 v2^2)

Ki = 0.5*(2*2*2 + 3*3*3)

Ki = 17.5

Kf = 0.5 (m1 + m2) v^2 = 0.5 * (2+3) * (1*1) = 2.5

energy = 17.5 - 2.5 = 15 J

c)

v1f = (m1-m2)/(m1+m2) v1i + (2m2)/(m1+m2) v2i

v1f = (2-3)/(2+3)* (-2) + (2*3)/(2+3) * (3) =

v1f = 4 m/s

v2f = 2m1/(m1+m2) v1i + (m2-m1)/(m1+m2) v2i

v2f = 2*2/(2+3)*(-2) + (3-2)/(2+3) * 3

v2f = -1 m/s

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