Question

Two blocks (m1 40 kg, m2 30 kg) are connected by a compressed spring, and are initially at rest. The spring breaks and the two objects fly apart from one another. If the speed of the first block is 15 m/s, what is the speed of the second block?
A. 15.0 m/s
B. 11.3 m/s
C.20.0 m/s
D.Cannot be determined

Answer 1

To develop this problem it is necessary to apply the equations related to the conservation of the moment, for which it is necessary that the velocities and the mass in an initial state must be equal to the mass and the velocity at the end point. Mathematically this can be described as

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

$m_1$ = Mass of object 1

$m_2$ = Mass of the object 2

$u_{1,2}$= Initital velocity of object 1 and 2

$v_{1,2}$ = Final velocity of object 1 and 2

Our values are given as,

$m_1 = 40kg$

$m_2 = 30kg$

There is not initial velocity, i.e, they are initially at rest.

We have the final velocity of the first object

$v_1 = 15m/s$

Replacing we have:

$m_1u_1+m_2u_2 = m_1v_1+m_2v_2$

$40*0+40*0=40*15+30*(v_2)$

$v_2=-20 m/s$

Therefore, the speed of the second block will be -20m / s or 20m / s if our vector reference system becomes positive for that direction. The correct answer is C.