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.

Accepted Answer

Step 1: Begin by applying the principle of conservation of momentum. Since the object was initially at rest, the total momentum before the explosion is zero. After the explosion, the momentum of the two fragments must still sum to zero. Let the mass of the lighter fragment be $$ m $$, and the mass of the heavier fragment be $$ 7m . If $$the velocities of the lighter and heavier fragments immediately after the explosion are $$ v_1 $$ and $$ v_2 $$, respectively, then $$ m v_1 = 7m v_2 . $$Solve for $$ v_1  in $$terms of $$ v_2 $$: $$ v_1 = 7v_2 . $$Step 2: Use the work-energy principle to relate the distance each fragment slides to its initial kinetic energy. The work done by friction is equal to the initial kinetic energy of each fragment. The work done by friction is $$ W = f_k d $$, where $$ f_k  is $$the kinetic friction force and $$ d  is $$the distance slid. The kinetic friction force is $$ f_k = \mu_k m g $$, where $$ \mu_k  is $$the coefficient of kinetic friction, $$ m  is $$the mass, and $$ g  is $$the acceleration due to gravity. Step 3: Write the kinetic energy for each fragment. The initial kinetic energy of the lighter fragment is $$ KE_1 = \frac{1}{2} m v_1^2 $$, and for the heavier fragment, $$ KE_2 = \frac{1}{2} (7m) v_2^2 . $$Substitute $$ v_1 = 7v_2 $$ into $$ KE_1 $$: $$ KE_1 = \frac{1}{2} m (7v_2)^2 = \frac{1}{2} m (49v_2^2) = 49 \cdot \frac{1}{2} m v_2^2 . $$Step 4: Relate the kinetic energy to the sliding distance for each fragment. For the heavier fragment, $$ KE_2 = f_k d_2 $$, where $$ d_2 = 8.2 \, 	ext{m} . $$For the lighter fragment, $$ KE_1 = f_k d_1 . $$Since $$ f_k  is $$proportional to mass, the ratio of distances $$ d_1 $$ and $$ d_2 $$ will depend on the ratio of their kinetic energies. Substitute $$ KE_1 $$ and $$ KE_2 $$ into the ratio: $$ \frac{d_1}{d_2} = \frac{KE_1}{KE_2} = \frac{49 \cdot \frac{1}{2} m v_2^2}{\frac{1}{2} (7m) v_2^2} = \frac{49}{7} = 7 . $$Step 5: Solve for $$ d_1 $$ using the ratio $$ \frac{d_1}{d_2} = 7 . $$Since $$ d_2 = 8.2 \, 	ext{m} $$, $$ d_1 = 7 \cdot d_2 = 7 \cdot 8.2 . $$The lighter fragment slides 7 times farther than the heavier fragment.

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.

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