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Chapter 7: Problem 54

A person stands in a stationary canoe and throws a \(5.00-\mathrm{kg}\) stone with a velocity of 8.00 \(\mathrm{m} / \mathrm{s}\) at an angle of \(30.0^{\circ}\) above the horizontal. The person and canoe have a combined mass of \(105 \mathrm{~kg}\). Ignoring air resistance and effects of the water, find the horizontal recoil velocity (magnitude and direction) of the canoe.

Short Answer

Expert verified
The horizontal recoil velocity of the canoe is approximately -0.377 m/s.

Step by step solution

01

Understand the Problem

A stone is thrown from a canoe, and we need to find the horizontal recoil velocity of the canoe due to the throw. Since no external horizontal forces are acting, the horizontal momentum of the system must be conserved.
02

Identify Given Values

The mass of the stone is given as 5.00 kg, the velocity of the stone is 8.00 m/s at an angle of 30.0 degrees above the horizontal, and the combined mass of the person and canoe is 105 kg.
03

Calculate Horizontal Velocity of the Stone

The horizontal component of the stone's velocity is given by \(v_{x} = v \cos(\theta)\), where \(v = 8.00\, \mathrm{m/s}\) is the velocity of the stone and \(\theta = 30.0^{\circ}\). Therefore, \(v_{x} = 8.00 \cos(30.0^{\circ})\).
04

Compute Horizontal Component of Stone's Momentum

The horizontal momentum of the stone is \(p_x = m_s \cdot v_x\), where \(m_s = 5.00\, \mathrm{kg}\). Substitute the value from the previous step: \(p_x = 5.00 \cdot (8.00 \cdot \cos(30.0^{\circ}))\).
05

Apply Conservation of Momentum

According to the conservation of momentum, the initial horizontal momentum (0, as everything is stationary) equals the final horizontal momentum. Thus, \(0 = -p_x + (m_c \cdot v_r)\), where \(m_c = 105\, \mathrm{kg}\) is the mass of the canoe and \(v_r\) is the recoil velocity of the canoe.
06

Solve for Canoe's Recoil Velocity

Rearrange the previous equation to find \(v_r\). We have \(v_r = \frac{p_x}{m_c}\). Substitute the value of \(p_x\) and \(m_c\) to calculate \(v_r\).
07

Calculate Numerical Result

Substitute the calculated momentum value to find \(v_r = \frac{5.00 \cdot 8.00 \cdot \cos(30.0^{\circ})}{105}\). Calculate this to get the final horizontal recoil velocity.

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Key Concepts

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

Momentum Conservation
Momentum conservation is a fundamental principle in physics. It states that the total momentum of a closed system remains constant when no external forces act upon it. In simpler terms, what goes in must come out, at least in terms of momentum.
  • In our exercise, the system consists of a person, a canoe, and a stone. Initially, everything is at rest, so the total momentum is zero.
  • When the stone is thrown, it gains momentum in the direction of its motion. To keep the total system momentum zero, the canoe must move in the opposite direction.
Think of it like a tiny explosion. You push the stone forward, and the canoe is pushed backward equally and oppositely. This backward motion of the canoe is what we call recoil.
Horizontal Velocity
Horizontal velocity refers to the component of velocity parallel to the ground. In projectile motion, it's essential to distinguish between the horizontal and vertical components.
  • When the stone is thrown at an angle of 30 degrees, its velocity has both horizontal and vertical parts.
  • To find just the horizontal velocity, trigonometry is used. The formula is:\[v_{x} = v \, \cos(\theta)\]where \(v\) is the stone's speed and \(\theta\) is the angle.
Grasping this distinction helps in understanding how the stone moves across the water versus how it rises and falls.
Recoil Velocity
Recoil velocity describes the speed and direction in which the canoe moves after the stone is thrown. It's akin to the kick you feel when firing a gun.
  • According to momentum conservation, the initial momentum (zero when everything is stationary) must be the same as the final momentum.
  • When solving for recoil velocity, one isolates the variable representing the canoe's velocity and rearranges the equation:
\[v_r = \frac{p_x}{m_c}\]where \(p_x\) is the horizontal momentum of the stone and \(m_c\) is the combined mass of the person and canoe.
This calculation reveals how fast and in what direction the canoe moves after the stone is thrown.
Vector Components
Vectors are quantities that have both magnitude and direction, making them suitable for describing velocities in physics. In our problem, the stone's velocity is a vector that can be divided into horizontal and vertical parts.
  • These parts are called components, derived using trigonometric functions based on the angle of the throw.
  • For instance, breaking down the stone's 8 m/s velocity at a 30-degree angle gives us a clearer understanding of how it affects the system's momentum.
By examining each component separately, we simplify complex scenarios into manageable calculations.
This approach helps make the effects of forces and motions in different directions more understandable.

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Most popular questions from this chapter

To view an interactive solution to a problem that is similar to this one, go to and select Interactive Solution \(\underline{7.55}\) A \(0.015-\mathrm{kg}\) bullet is fired straight up at a falling wooden block that has a mass of \(1.8 \mathrm{~kg}\). The bullet has a speed of \(810 \mathrm{~m} / \mathrm{s}\) when it strikes the block. The block originally was dropped from rest from the top of a building and had been falling for a time \(t\) when the collision with the bullet occurred. As a result of the collision, the block (with the bullet in it) reverses direction, rises, and comes to a momentary halt at the top of the building. Find the time \(t\).

ssm Batman (mass \(=91 \mathrm{~kg}\) ) jumps straight down from a bridge into a boat (mass \(=510 \mathrm{~kg}\) ) in which a criminal is fleeing. The velocity of the boat is initially \(+11\) \(\mathrm{m} / \mathrm{s}\). What is the velocity of the boat after Batman lands in it?

A ball is dropped from rest at the top of a \(6.10\) -m-tall building, falls straight downward. collides inelastically with the ground, and bounces back. The ball loses \(10.0 \%\) of its kinetic energy every time it collides with the ground. How many bounces can the ball make and still reach a windowsill that is \(2.44 \mathrm{~m}\) above the ground?

A golfer, driving a golf ball off the tee, gives the ball a velocity of \(+38 \mathrm{~m} / \mathrm{s}\). The mass of the ball is \(0.045 \mathrm{~kg},\) and the duration of the impact with the golf club is \(3.0 \times 10^{-3} \mathrm{~s}\). (a) What is the change in momentum of the ball? (b) Determine the average force applied to the ball by the club.

At in preparation for this problem. Two friends, Al and Jo, have a combined mass of \(168 \mathrm{~kg}\). At an ice skating rink they stand close together on skates, at rest and facing each other, with a compressed spring between them. The spring is kept from pushing them apart because they are holding each other. When they release their arms, Al moves off in one direction at a speed of \(0.90 \mathrm{~m} / \mathrm{s}\), while Jo moves off in the opposite direction at a speed of \(1.2 \mathrm{~m} / \mathrm{s}\). Assuming that friction is negligible, find Al's mass.
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