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A professor of physics is going ice skating for the first time. He has gotten himself into the middle of an ice rink and cannot figure out how to make the skates work. Every motion he makes simply causes his feet to slip on the ice and leaves him in the same place he started. He decides that he can get off the ice by throwing his gloves in the opposite direction. (a) Suppose he has a mass \(M\) and his gloves have a mass \(m\). If he throws the gloves as hard as he can away from him, they leave his hand with a velocity \(\vec{v}_{\text {glove }}\) Explain whether or not he will move. If he does move, calculate his velocity, \(\vec{v}_{\text {prof }}\) (b) Discuss his motion from the point of view of the forces acting on him. (c) If the ice rink is \(10 \mathrm{~m}\) in diameter and the skater starts in the center, estimate how long it will take him to reach the edge, assuming there is no friction at all.

Step by step solution

01

Analyze the situation

The professor initially is at rest in the middle of the ice rink. When he throws the gloves, we need to consider the law of conservation of momentum to understand his motion.

02

Use the Conservation of Momentum (Part a)

The total momentum before he throws the gloves is zero because both the professor and the gloves are stationary. After he throws the gloves, the total momentum must still be zero. Thus, the momentum of the professor and the gloves must be equal and opposite. Mathematically: \[\text{Total Initial Momentum} = 0 = M \times \text{Initial Velocity of Professor} + m \times \text{Initial Velocity of Gloves}\] Since both are initially at rest, zero momentum initially will be: \[0 = M \times \text{v}_{\text{prof}} + m \times \text{v}_{\text{glove}}\] Solving for the professor's velocity, \[\text{v}_{\text{prof}} = - \frac{m \times \text{v}_{\text{glove}}}{M}\]

03

Determine the Professor's Velocity (Part a)

Using the equation derived in Step 2, calculate the professor's velocity. \[\text{v}_{\text{prof}} = - \frac{m \times \text{v}_{\text{glove}}}{M}\]

04

Discuss Forces Acting on Professor (Part b)

While on the ice, the professor needs to exert a force on the gloves to throw them. By Newton's third law, there is an equal and opposite force exerted on the professor by the gloves. This force causes the professor to move in the opposite direction.

05

Calculate Time to Reach Edge of Rink (Part c)

The professor starts from the center of a rink with a diameter of 10 meters, so he needs to travel 5 meters (the radius). Using the velocity from Step 3, the required time is:\[\text{Time} = \frac{\text{Distance to edge}}{\text{Velocity}} = \frac{5 m}{\text{v}_{\text{prof}}} \]Plugging in the value for \text{v}_{\text{prof}},\ the expression becomes:\[\text{Time} = \frac{5}{- \frac{m \times \text{v}_{\text{glove}}}{M}} = \frac{5M}{m \times \text{v}_{\text{glove}}} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's third law

Newton's third law states that for every action, there is an equal and opposite reaction. This principle is crucial in understanding how the professor moves when he throws his gloves. When the professor exerts a force to throw his gloves in one direction, the gloves exert an equal and opposite force on him. This opposite force propels him in the direction opposite to that in which he threw the gloves. Because of this law, any movement the professor makes has a direct and immediate impact on his motion.

Key takeaways:

• The force exerted on the gloves is equal in magnitude but opposite in direction to the force exerted on the professor.
• The direction of the professor's movement is opposite to that of the gloves.

Understanding Newton's third law helps us see why he won't stay stationary after throwing his gloves.

frictionless surface

The scenario takes place on an ice rink, assumed to be a frictionless surface. Friction is the resistance that one surface or object encounters when moving over another. In this case, the ice provides almost no resistance to the professor's movements.

Important aspects of a frictionless surface:

• There is no frictional force to oppose the professor's movement once the gloves are thrown.
• Any small force can set the professor in motion because there’s no friction to counteract it.

The absence of friction ensures that all forces acting are purely the result of the throw, making the calculations simpler. Hence, the velocity calculated is a straightforward result of the conservation of momentum alone, without adjustments for friction.

velocity calculation

The conservation of momentum is essential in calculating the professor's velocity after he throws the gloves. Since he and the gloves are initially at rest, their total momentum is zero. The professor's velocity can be determined using the equation: \[ \text{v}_{\text{prof}} = - \frac{m \times \text{v}_{\text{glove}}}{M} \]
Steps involved:

• Identify that initial momentum is zero because both are initially stationary.
• Use the momentum conservation formula: \[ 0 = M \times \text{v}_{\text{prof}} + m \times \text{v}_{\text{glove}} \]
• Rearrange the equation to solve for \[ \text{v}_{\text{prof}} \]

This formula tells us that the professor's velocity is inversely proportional to his mass and directly proportional to the velocity at which he throws the gloves and their mass.
If the professor has a larger mass, his resulting velocity will be smaller for the same gloves' mass and velocity. Hence, the professor's mass and the manner of throwing the gloves directly determine how quickly he will glide across the ice.