A 07 kg mass particle m is shot upward at 13 ms into the tub

A 0.7 kg mass particle m is shot upward at 13 m/s into the tube as shown. The surface of the tube is frictionless. r = 0.5 m. What is the velocity with which the mass arrives at the top of the tube? m/s

Solution

Given

mass of the particle, m = 0.7 kg

Velocity at point A, v_A = 13 m/s

Total energy of the system (particle + tube) at point A

E_A = Kinetic energy at point A + Potential energy at point A

The energy components of the tube are zero since the tube is always at rest

E_A = 0.5mv_A² + mgh_A

where h_A is the height of the particle from the ground at point A

Similarly

Total energy of the system at point B

E_A = 0.5mv_B² + mgh_B

Since there is no friction involved, hence the total energy is conserved

E_A = E_B

0.5mv_A² + mgh_A = 0.5mv_B² + mgh_B

Dividing by \'0.5m\' on both the sides

v_B² = 2g(h_A - h_B) + v_A²

= 2 X 9.81 X (-6) + 13²

= 51.28

Applying square root on both sides

v_B = 7.161 m/s..................................(Answer)