Problem 20
Question
You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a 0.400 -kg ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move afterward? (b) If the ball hits you and bounces off your chest, so afterward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?
Step-by-Step Solution
Verified Answer
(a) 0.0568 m/s; (b) 0.103 m/s.
1Step 1: Understand the Problem
We're given a situation where we need to determine the speed of you and a ball after a collision on a frictionless surface. We need to solve two parts: (a) when you catch the ball, and (b) when the ball bounces back.
2Step 2: Conservation of Momentum for (a)
For part (a), we apply conservation of momentum. Initially, the total momentum is due to the ball: \[ p_{initial} = m_{ball} \times v_{ball} = 0.400 \times 10.0 = 4.0 \text{ kg m/s} \]After catching, you and the ball move together, so:\[ p_{final} = (m_{you} + m_{ball}) \times v_{final} \]Equating initial and final momentum gives:\[ 4.0 = (70.0 + 0.400) \times v_{final} \]
3Step 3: Solve for Final Velocity in (a)
Solve for \(v_{final}\):\[ 4.0 = 70.4 \times v_{final} \]\[ v_{final} = \frac{4.0}{70.4} \approx 0.0568 \text{ m/s} \]
4Step 4: Conservation of Momentum for (b)
For part (b), again apply conservation of momentum. Initially:\[ p_{initial} = 4.0 \text{ kg m/s} \] The ball bounces back and a new momentum for the ball is:\[ p'_{ball} = m_{ball} \times (-8.0) = 0.400 \times (-8.0) = -3.2 \text{ kg m/s} \] After collision, the total momentum is:\[ p_{final} = m_{you} \times v_{you} + p'_{ball} \]Equating initial and final momentum:\[ 4.0 = 70.0 \times v_{you} - 3.2 \]
5Step 5: Solve for Your Velocity in (b)
Rearrange and solve for \(v_{you}\):\[ 4.0 + 3.2 = 70.0 \times v_{you} \]\[ 7.2 = 70.0 \times v_{you} \]\[ v_{you} = \frac{7.2}{70.0} \approx 0.103 \text{ m/s} \]
Key Concepts
Collision PhysicsFrictionless SurfaceMomentum Calculation
Collision Physics
In collision physics, we are interested in what happens when two objects hit each other. Collisions can happen in various ways, but they mostly adhere to the same basic principles. These principles involve changes in the velocity and direction of the involved objects and are governed mainly by the conservation of momentum.
In any collision, we consider:
- The initial velocities of the objects before they collide.
- The final velocities of the objects after the collision.
- Whether the collision is elastic or inelastic, which will tell us whether kinetic energy is conserved along with momentum.
Frictionless Surface
A frictionless surface is an ideal concept used often in physics problems to simplify the calculations. In reality, most surfaces have some degree of friction, but by assuming frictionlessness, we can better understand the core interactions between objects without the additional complexity presented by frictional forces.
When there is no friction:
- Moving objects do not lose energy to the surface they are on.
- There is no resistance to the change in motion or direction.
- The only forces you need to consider are those directly acting on the objects (such as the force from a collision).
Momentum Calculation
Momentum is one of the core concepts in physics and is defined as the product of an object's mass and its velocity. It is a vector quantity, possessing both magnitude and direction, and is vital when calculating the outcomes of collisions. The formula for momentum is:\[p = m imes v\]Where:
- \( p \) is momentum.
- \( m \) is the mass of the object.
- \( v \) is the velocity of the object.
Other exercises in this chapter
Problem 18
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On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass \(0.350 \mathrm{kg} ),\) which is in
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