Question

Physics question WILL GIVE BRAINLIEST!!! A 1,600 kg train car rolling freely on level track at 16 m/s bumps into a 1.0 x 10^3 kg train car moving at 10.0 m/s in the same direction, and the two latch onto each other and continue together. What is their final speed?
10 m/s
14 m/s
6.0 m/s
16 m/s

Answer 1

Answer:

$v'\approx 14\ m/s$

Explanation:

Linear Momentum

Two objects of masses and moving in a linear path at speeds and respectively have a total momentum of

$p=m_1v_1+m_2v_2$

When the objects collide, a change of conditions occurs and they start to move at different speeds. The necessary condition to find the after-colliding speeds is the conservation of linear momentum that states the total momentum of an isolated system doesn't change regardless of the internal interactions of the objects. Thus, the new momentum is

$p'=m_1v_1'+m_2v_2'$

And they must be the same, thus

$m_1v_1'+m_2v_2'=m_1v_1+m_2v_2$

We know both cars stick together after the collision, so the final speed is common to both, and the above formula becomes

$(m_1+m_2)v'=m_1v_1+m_2v_2$

Solving for v'

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{(m_1+m_2)}$

Plugging in the values, we have

$\displaystyle v'=\frac{(1600)(16)+(1000)(10)}{(1600+1000)}$

$\boxed{v'=13.69\ m/s}$

Correct option (closest to the computed speed): 14 m/s