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.

Verified step by step guidance

1. 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 \( v_2 \): \( v_1 = 7v_2 \).
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.
3. 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 \( KE_1 \): \( KE_1 = \frac{1}{2} m (7v_2)^2 = \frac{1}{2} m (49v_2^2) = 49 \cdot \frac{1}{2} m v_2^2 \).
4. Step 4: Relate the kinetic energy to the sliding distance for each fragment. For the heavier fragment, \( KE_2 = f_k d_2 \), where \( d_2 = 8.2 \, \text{m} \). 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: \( \frac{d_1}{d_2} = \frac{KE_1}{KE_2} = \frac{49 \cdot \frac{1}{2} m v_2^2}{\frac{1}{2} (7m) v_2^2} = \frac{49}{7} = 7 \).
5. Step 5: Solve for \( d_1 \) using the ratio \( \frac{d_1}{d_2} = 7 \). Since \( d_2 = 8.2 \, \text{m} \), \( d_1 = 7 \cdot d_2 = 7 \cdot 8.2 \). The lighter fragment slides 7 times farther than the heavier fragment.

Key Concepts

Conservation of Momentum

The principle of conservation of momentum states that in a closed system, the total momentum before an event must equal the total momentum after the event. In this scenario, since the object is initially at rest, the total momentum is zero. After the explosion, the momentum of the two fragments must also sum to zero, allowing us to relate their masses and velocities.

Friction and Kinetic Friction Coefficient

Friction is the force that opposes the relative motion of two surfaces in contact. The coefficient of kinetic friction quantifies this force and is constant for a given pair of surfaces. In this problem, both fragments experience the same frictional force, which affects how far they slide after the explosion, allowing us to compare their distances based on their masses.

Kinematic Equations

Kinematic equations describe the motion of objects under constant acceleration. In this case, the acceleration is due to friction acting on the fragments after the explosion. By applying these equations, we can determine the distance each fragment travels based on its initial velocity and the deceleration caused by friction.