Question

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavier fragment slides $8.2$m before stopping. How far does the lighter fragment slide? Assume that both fragments have the same coefficient of kinetic friction

Short answer

The distance traveled by the lighter fragment side is$400m$

Step-by-step solution

Step 1. Given information

One fragment is seven times as massive as the other fragment.

The distance traveled by heavier fragment,$8.2m$

Step 2. Evaluating the velocities

Use the Newton's second law of motion,

$\begin{array}{c}{F}_{\text{heavier}}=a_{\text{heavier}}{m}_{\text{heavier}}={\mu }_{k}g\\ F_{\text{lighter}}=a_{\text{lighter}}{m}_{\text{lighter}}={\mu }_{k}g\end{array}$

Now, evaluating the velocities,

For heavier:

$\begin{array}{r}{v}_{\text{heavier}}^2=2d_{\text{heavier}}\\ v_{\text{heavier}}=\sqrt{2\frac{{\mu }_{k}g}{m_{\text{heavier}}}{d}_{\text{heavier}}}\end{array}$

For lighter:

$\begin{array}{r}{v}_{\text{lighter}}^2=2a_{\text{lighter}}{d}_{\text{lighter}}\\ v_{\text{lighter}}=\sqrt{2\frac{{\mu }_{k}g}{m_{\text{lighter}}}{d}_{\text{lighter}}}\end{array}$

$\begin{array}{l}\frac{{v_{lighter}}^2}{{v_{heavier}}^2}=\frac{d_{lighter}}{d_{heavier}}\\ \Rightarrow \frac{d_{lighter}}{d_{heavier}}=49\\ \Rightarrow d_{lighter}=49d_{heavier}\end{array}$

Step 3. Applying conservation principle

On applying conservation of momentum we have,

$$\begin{gathered} 0=m_{heavier}{v}_{heavier}-m_{lighter}{v}_{lighter} \\ {m}_{heavier}\sqrt{2\frac{{\mu }_{k}g}{m_{heavier}}{d}_{heavier}}-m_{heavier}\sqrt{2\frac{{\mu }_{k}g}{m_{heavier}}{d}_{lighter}}=0 \\ {d}_{lighter}=49\times 8.2\approx 400m \end{gathered}$$