Question

Two hockey pucks on an ice rink are held together with a compressed spring between them. The pucks are released and the spring pushes them in opposite directions. One puck of mass 0.5 kg moves at 8 m/s. Calculate the speed of the other puck of mass 2 kg

Answer 1

Answer:

$v_{2f} = -2 m/s$

so the puck of 2 kg will move in opposite direction with speed 2 m/s

Explanation:

Since two pucks are compressed by the spring between them so there is no external force on this system

so here momentum of the system of two pucks will remains conserved

here we can have

$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$

here initially both pucks are at rest

so we can have

$0 + 0 = 0.5(8 m/s) + 2(v_{2f})$

now the final speed of 2kg is given as

$v_{2f} = - 2 m/s$

Answer 2

Where the m1 and m2 is the mass of two object. U1 and U2 are the initial velocity of hockey pucks. V1 and V2 are the final velocity of Hockey Pucks.

m2=2kg The initial velocity of both pucks are 0. The v1=8m/s. All you have to do now is find v2

0=4-2v
-4=-2v
v=+2m/s

Answer: +2