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At what speed do a bicycle and its rider, with a combined mass of \(100 \mathrm{kg}\), have the same momentum as a \(1500 \mathrm{kg}\) car traveling at \(5.0 \mathrm{m} / \mathrm{s} ?\)

Short Answer

Expert verified
The bicycle and the rider, with a combined mass of 100 kg, must be traveling at a speed of 75 m/s to have the same momentum as a 1500 kg car traveling at 5.0 m/s.

Step by step solution

01

Calculate the Momentum of the Car

Calculate the momentum of the car using the formula for momentum, \( p = mv \), where \( m \) is the mass and \( v \) is the velocity of the car. Given in the problem, the mass of the car is 1500 kg and its velocity is 5.0 m/s. Input these values into the formula to get the momentum: \( p = (1500 \, \mathrm{kg})(5.0 \, \mathrm{m/s}) = 7500 \, \mathrm{kg m/s} \)
02

Calculate the Velocity of the Bike

Use the momentum of the car calculated in the previous step and the given mass of the bike and its rider to calculate their velocity. Use the formula for momentum again, this time solving for \( v \): \( v = p/m \). The given mass of the bike and its rider is 100 kg and the momentum we need is 7500 kg m/s. Input these values into the formula to find the velocity: \( v = 7500 \, \mathrm{kg m/s} / 100 \, \mathrm{kg} = 75 \, \mathrm{m/s} \)

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Key Concepts

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

Momentum Formula
Momentum in physics can be considered as a measure of how difficult it is to stop a moving object. It is a vector quantity, meaning it has both magnitude and direction. The momentum formula plays an essential role in understanding the movement of objects and is a foundational concept in both classical and modern physics. The basic momentum formula is expressed as:\[ p = mv \]
where \( p \) stands for momentum, \( m \) is the mass of the object, and \( v \) is its velocity. The product of mass and velocity gives us an object's linear momentum, which is conserved in closed systems according to the conservation laws of physics. In our example, we calculated the momentum of a car and then used that information to find the required speed of a bicycle for it to have equal momentum. Understanding how to manipulate this formula is crucial in solving many problems related to object dynamics.
Linear Momentum
Linear momentum, or simply momentum, is the quantity of motion of a moving body. It is directly proportional to both the object’s mass and its velocity, making it an extensive property: the greater the mass or velocity, the greater the momentum. This is important when considering objects in motion as it helps predict their subsequent movements when interacting with other objects or forces.
For instance, to have the same momentum as a heavy, slow-moving vehicle, a lighter one must move quite fast. As demonstrated in the exercise, to match a car's momentum, the velocity of a bicycle has to be significantly higher due to its lower mass. This highlights a crucial aspect of linear momentum: it serves as a bridge between the motion (kinematics) and the causes of motion (dynamics) in physics.
Conservation of Momentum
The conservation of momentum is a principle stating that the total momentum of a closed system remains constant if no external forces are applied. This fundamental law of physics is derived from Newton's third law, which states that for every action, there is an equal and opposite reaction. In closed or isolated systems, momentum exchanges between objects without any loss, describing predictable outcomes of collisions and interactions.
In the context of our bicycle and car example, if the two were to collide in an ideal situation without external influences, the total momentum just before impact would equal the total momentum after impact. This law helps us understand systems ranging from simple physics exercises to complex astronomical events.

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Most popular questions from this chapter

A firecracker in a coconut blows the coconut into three picces. Two pieces of equal mass fly off south and west, perpendicular to each other, at \(20 \mathrm{m} / \mathrm{s}\). The third piece has twice the mass as the other two. What are the speed and direction of the third piece? Give the direction as an angle east of north.

A 10 -m-Iong glider with a mass of \(680 \mathrm{kg}\) (including the passengers) is gliding horizontally through the air at \(30 \mathrm{m} / \mathrm{s}\) when a \(60 \mathrm{kg}\) skydiver drops out by releasing his grip on the glider. What is the glider's velocity just after the skydiver lets go?

A \(20 \mathrm{kg}\) wood ball hangs from a 2.0 -m-long wire. The maximum tension the wire can withstand without breaking is 400 N. A \(1.0 \mathrm{kg}\) projectile traveling horizontally hits and embeds itself in the wood ball. What is the largest speed this projectile can have without causing the cable to break?

A 20 g ball of clay is shot to the right (in the positive \(x\) direction) at \(12 \mathrm{m} / \mathrm{s}\) toward a \(40 \mathrm{g}\) ball of clay at rest. The two balls of clay collide and stick together. Call this reference frame \(S\). a. What is the total momentum in frame S? b. What is the velocity \(\vec{V}\) of a reference frame \(S^{\prime}\) in which the total momentum is zero? c. After the collision, what is the velocity in frame S' of the resulting \(60 \mathrm{g}\) ball of clay? Answering this question requires only thought, no calculations. d. Use your answer to part \(c\) and the Galilean transformation of velocity to find the post-collision velocity of the \(60 \mathrm{g}\) ball of clay in reference frame S.

A two-stage rocket is traveling at \(1200 \mathrm{m} / \mathrm{s}\) with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of \(35 \mathrm{m} / \mathrm{s}\) relative to the second stage. The first stage is three times as massive as the second stage. What is the speed of the second stage after the separation?

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