Question

Block A begins at the top of the track with an initial speed. It then slides on the track without friction going down to the bottom and back up to the second platform. Finally, it collides with a stationary block, B and the two stick together and continue to slide.

Block A, with mass, mA= 0.75 kg is given an initial velocity, v0= 1.3 m/s. Block A slides without friction along a track downward from a platform at height h1= 0.80 m to ground level and back upward to a platform at level h2= 0.25 m. It then collides with and sticks to a stationary block B, with mass, mB= 0.3 kg. What is the final velocity of the two blocks after the collision? Answer in m/s.

Answer 1

By energy conservation between initial position and position just before collision

(Kinetic Energy + potential Energy)initial =( K. E + P. E)final ​​​​​​

1/2 mv12+ mgh1 = 1/2 mv22 + mgh2​​​​​

1/2 (1.3)2 + 10 × 0.8 = 1/2 v22 + 10×0.25

v2 = 3.56 m/s

For collision

By momentum conservation

mAv2 + 0 = (mA+mB​​​​​​) V

0.75 × 3.56 = (0.75+0.3) V

V = 2.543 m/s