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Chapter 9: Problem 58

\(A\) 20 \(g\) ball of clay traveling east at \(2.0 \mathrm{m} / \mathrm{s}\) collides with a \(30 \mathrm{g}\) ball of clay traveling \(30^{\circ}\) south of west at \(1.0 \mathrm{m} / \mathrm{s}\). What are the speed and direction of the resulting \(50 \mathrm{g}\) blob of clay?

Short Answer

Expert verified
The resulting blob of clay moves towards \(46.2^{\circ}\) south of east with a speed of \(1.037 m/s\).

Step by step solution

01

Calculate initial momenta

The initial momentum is given by mass times velocity. For the first clay, momentum \(p_{1} = m_{1} \cdot v_{1}\) yields \(p_{1} = 20 \cdot 2 = 40 \mathrm{g \, m/s}\) East. For the second clay, momentum \(p_{2} = m_{2} \cdot v_{2}\) yields \(p_{2} = 30 \cdot 1 = 30 \mathrm{g \, m/s}\) at \(30^{\circ}\) south of west.
02

Convert into Cartesian coordinates

To add the two vectors, we need to convert them into Cartesian coordinates. For \(p_{1}\), the x-component is 40 (since it鈥檚 going eastonly, no y-component). For \(p_{2}\), we project the vector onto the axis: \(p_{2_{x}} = p_{2} \cdot \cos(120^{\circ}) = -15 \, \mathrm{g \, m/s}\); \(p_{2_{y}} = p_{2} \cdot \sin(120^{\circ}) = -25.98 \, \mathrm{g \, m/s}\).
03

Find total momentum

We can add the two momenta vectorially by adding the i-components and the j-components separately. Total momentum \(\vec{P}_{T} = \vec{p}_{1} + \vec{p}_{2}\) which gives \(\vec{P}_{T_{x}} = p_{1_{x}} + p_{2_{x}} = 40 - 15 = 25 \, \mathrm{g \, m/s}\), and \(\vec{P}_{T_{y}} = p_{1_{y}} + p_{2_{y}} = 0 - 25.98 = -25.98 \, \mathrm{g \, m/s}\).
04

Calculate speed and direction

The magnitude of the total momentum divided by the total mass will give the speed: \(v_{T} = |\vec{P}_{T}| / m_{T} = \sqrt{(25^2 + (-25.98)^2)} / 50 = 1.037 \, \mathrm{m/s}\). For the direction, take the arctan of the y-component divided by the x-component: \(\theta = \arctan (\frac{P_{T_{y}}}{P_{T_{x}}}) = \arctan (\frac{-25.98}{25}) = -46.2^{\circ}\). Because it is in the 4th quadrant, the actual angle is \(\theta + 360^{\circ} = 313.8^{\circ}\) which is \(46.2^{\circ}\) south of east.

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Key Concepts

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

Collision Analysis
When two objects collide, like the balls of clay in this exercise, we analyze the impact by examining their respective speeds, directions, and masses. This is key to determining the aftermath of the collision.

For our clay balls, the collision is perfectly inelastic, meaning they stick together post-collision. Understanding such scenarios helps us conserve momentum to predict the resulting motion.

We'll need their initial directions and velocities. One ball is moving east at 2.0 m/s, while the other is moving southwest at a 30掳 angle with a speed of 1.0 m/s. These vectors will guide our calculation of the total momentum post-collision.
Vector Addition
Adding vectors involves breaking them into components, especially when they're not aligned. Each component represents a part of the vector along an axis鈥攈orizontal (x-axis) and vertical (y-axis).

For our scenario, the first ball has momentum only along the x-axis, since it's traveling east.
  • Momentum (p) = mass (m) 脳 velocity (v)
  • For the eastward clay: 20 g 脳 2 m/s = 40 g m/s in the x-direction
The second clay鈥檚 momentum involves trigonometric functions to find direction along both axes. Applying cosine and sine helps us determine:
  • Horizontal component: \[ p_{2_{x}} = p_{2} \times \cos(120^{\circ}) = -15 \text{ g m/s} \]
  • Vertical component: \[ p_{2_{y}} = p_{2} \times \sin(120^{\circ}) = -25.98 \text{ g m/s} \]
After breaking them down, we add these components to find the total momentum in each direction.
Momentum Calculation
Momentum describes the quantity of motion. It's crucial to calculate each object's momentum before adding them to find the total.

The total momentum comes from adding up the separate x and y components.

For the x-components:
  • \[ P_{T_{x}} = p_{1_{x}} + p_{2_{x}} = 40 - 15 = 25 \text{ g m/s}\]
For the y-components:
  • \[ P_{T_{y}} = p_{1_{y}} + p_{2_{y}} = 0 - 25.98 = -25.98 \text{ g m/s} \]
The resulting vector \( \vec{P}_T \) is found combining these components. Magnitude of the vector gives the blob's speed: \[ v_{T} = \frac{\sqrt{(25^2 + (-25.98)^2)}}{50} = 1.037 \text{ m/s} \] Thus, we calculate how fast the two pieces move together after the collision.
Direction Determination
Determining direction involves using inverse trigonometric functions. Here, we focus on the arctangent to find the angle of the momentum vector.

The arctangent function relates the y-component to the x-component:
  • \[ \theta = \arctan \left(\frac{-25.98}{25}\right) = -46.2^{\circ} \]
The angle is negative, placing it in the 4th quadrant of a coordinate plane. We adjust this angle to more standard geography:
  • Adding 360掳 to -46.2掳 yields 313.8掳鈥攎aking it 46.2掳 south of east.
This tells us exactly how the combined mass has changed direction, enabling us to visualize the resultant path.

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Most popular questions from this chapter

Ann (mass \(50 \mathrm{kg}\) ) is standing at the left end of a 15 -m-long. 500 kg cart that has frictionless wheels and rolls on a frictionless track. Initially both Ann and the cart are at rest. Suddenly, Ann starts running along the cart at a speed of \(5.0 \mathrm{m} / \mathrm{s}\) relative to the cart. How far will Ann have run relative to the ground when she reaches the right end of the cart?

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 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.

Force \(F_{x}=(10 \mathrm{N}) \sin (2 \pi t / 4.0 \mathrm{s})\) is exerted on a \(250 \mathrm{g}\) parti- cle during the interval 0 s \(\leq t \leq 2.0\) s. If the particle starts from rest, what is its speed at \(t=2.0 \mathrm{s} ?\)

In Problem 27 of Chapter 7 you found the recoil speed of Bob as he throws a rock while standing on frictionless ice. Bob has a mass of \(75 \mathrm{kg}\) and can throw a \(500 \mathrm{g}\) rock with a speed of \(30 \mathrm{m} / \mathrm{s}\) Find Bob's recoil speed again, this time using momentum.
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