Question

Three carts move on a frictionless track with inertia

Answer 1

Answer:

Given: The mass of the first cart {eq}m_1=4.82 \ \text{kg} {/eq} which has the velocity {eq}u_1=+5.03 \ \text{m/s} {/eq}

The mass of the second cart {eq}m_2=2.72 \ \text{kg} {/eq} which has the velocity {eq}u_2=-3.52 \ \text{m/s} {/eq}

They have skidded off together after the collision suppose the final velocity of the combined carts is {eq}v {/eq}.

From the momentum conservation:

{eq}m_1u_1+m_2u_2=(m_1+m_2)v {/eq}

Thus, the velocity of the combined bodies can be computed as:

{eq}\begin{align} v&=\frac{(m_1u_1+m_2u_2)}{(m_1+m_2)}\\ v&=\frac{(4.82\times5.03+2.72\times(-3.52))}{(4.82+2.72)}\\ v&=\color{blue}{1.946 \ \text{m/s}}\\ \end{align} {/eq}