Problem 60 - College Physics

Question

\(\bullet\) A 5.00 kg ornament is hanging by a 1.50 \(\mathrm{m}\) wire when it is suddenly hit by a 3.00 \(\mathrm{kg}\) missile traveling horizontally at 12.0 \(\mathrm{m} / \mathrm{s} .\) The missile embeds itself in the ornament during the collision. What is the tension in the wire immediately after the collision?

Step-by-Step Solution

Answer

The tension in the wire is 186.4 N.

Step 1: Understand the Problem

We have a 5.00 kg ornament hanging by a wire. A 3.00 kg missile moving horizontally at 12.0 m/s strikes the ornament and becomes embedded in it. We need to find the tension in the wire immediately after the collision.

Step 2: Calculate the Combined Mass

When the missile embeds into the ornament, their masses combine for the purposes of calculating after-collision properties.The total combined mass is:\[m_{total} = m_{ornament} + m_{missile} = 5.00 \, \text{kg} + 3.00 \, \text{kg} = 8.00 \, \text{kg}\]

Step 3: Use the Principle of Conservation of Momentum

The law of conservation of momentum states that the total momentum before an event must equal the total momentum after the event. Before the collision, the momentum of the system is given by the momentum of the missile since the ornament is stationary:\[p_{initial} = m_{missile} \times v_{missile} = 3.00 \, \text{kg} \times 12.0 \, \text{m/s} = 36.0 \, \text{kg m/s}\]

Step 4: Calculate Velocity After Collision

After the collision, we assume the system moves horizontally with a velocity \(v_{final}\)Using conservation of momentum:\[p_{initial} = p_{final} \rightarrow 36.0 \, \text{kg m/s} = 8.00 \, \text{kg} \times v_{final}\]Solve for \(v_{final}\):\[v_{final} = \frac{36.0}{8.00} = 4.50 \, \text{m/s}\]

Step 5: Calculate the Tension in the Wire

Immediately after the collision, the ornament-missile system will experience a centripetal force as it moves downward in a circular trajectory due to gravity.The tension \(T\) in the wire can be determined by the net forces in the vertical direction. This includes the gravitational force and the centripetal force required to maintain the curvature of the path:\[T = m_{total} \times g + \frac{m_{total} \times (v_{final})^2}{L}\]where:- \(g = 9.8 \, \text{m/s}^2\) (acceleration due to gravity)- \(L = 1.50 \, \text{m}\) (length of the wire)First, compute the gravitational force:\[F_{gravity} = 8.00 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 78.4 \, \text{N}\]Second, compute the centripetal force:\[F_{centripetal} = \frac{8.00 \, \text{kg} \times (4.50 \, \text{m/s})^2}{1.50 \, \text{m}} = 108.0 \, \text{N}\]Finally, calculate the tension:\[T = 78.4 \, \text{N} + 108.0 \, \text{N} = 186.4 \, \text{N}\]

Step 6: Provide the Final Answer

The tension in the wire immediately after the collision is calculated to be 186.4 N.

Key Concepts

Centripetal Force

Centripetal force is the force that keeps an object moving in a circular path. It constantly acts towards the center of the circle. Whether you are swinging a ball on a string, or driving a car around a circular track, centripetal force is at play.

In the context of our exercise, when the 8 kg ornament-missile system swings in a circular path, a centripetal force is necessary to maintain this motion. Without this force, the object would fly off in a straight line due to inertia. The centripetal force here is supplied by the tension in the wire.

The formula for centripetal force is:\[F_{centripetal} = \frac{m imes v^2}{r}\]where:

• \( m \) is the mass of the object, in this case, 8 kg.
• \( v \) is the velocity, found to be 4.5 m/s post-collision.
• \( r \) is the radius, equivalent to the wire length, 1.5 m.

If any one of these factors changes, the centripetal force will adjust to maintain its path. Understanding this balance assists in solving how tension affects the centripetal motion in these scenarios.

Tension in Wire

Tension in a wire or string is the force exerted along the length, which supports the suspended object's weight and any added forces due to motion. In physics problems involving objects in motion, calculating the wire tension is crucial for understanding how the forces interact.

In our exercise, after the missile-embedment collision, the ornament begins to move in a circular path. The tension must counteract gravitational force while providing the necessary centripetal force for circular motion. Thus, the wire experiences tension from two main sources:

• **Gravitational Force**: This is simply the weight of the combined mass. Here, it's calculated as \(78.4 \: N\).
• **Centripetal Force**: Necessary to maintain the circular path, calculated to be \(108.0 \: N\).

The total tension in the wire, immediately after collision, is the sum of these forces:\[T = F_{gravity} + F_{centripetal} = 78.4 \: N + 108.0 \: N = 186.4 \: N\]By understanding each component's role in the overall tension, we see how they combine to sustain the system's dynamic equilibrium.

Collision Physics

In the realm of physics, collisions are events where two or more bodies exert forces on each other in a relatively short time period. They follow the principles of conservation laws, namely momentum and, sometimes, energy, depending on the collision type.

Conservation of momentum is a key aspect here. Before the collision, the missile has a momentum of \(36.0 \: \text{kg m/s}\); the stationary ornament has none. After the collision, the combined system (ornament plus embedded missile) must also have a momentum of \(36.0 \: \text{kg m/s}\). This allows for the calculation of the system's velocity post-collision.

However, inelastic collisions, like the one described where objects merge, often involve kinetic energy loss due to factors like deformation or sound. Here, focus remains solely on momentum conservation due to its straightforward implications for immediate velocities. Note that understanding the conservation of momentum principle is crucial for solving complex collision problems accurately and deciphering the sequence of forces in action.