https://r2.erweima.ai/mathbot/image/1726395411302c3y0r6z0.jpg

Let's begin by analyzing and solving the problems in the image:

### **Bài 1:**
This problem involves the application of force on a mass of 2.5 kg over time, as shown by the graph. We will answer the parts one by one.

#### (a) Impulse Calculation from 0 to 5 seconds:
The impulse \( J \) is given by:
\[
J = \int F(t) \, dt
\]
This is the area under the force-time graph. We will calculate the total area between 0 to 5 seconds (this area is trapezoidal).

#### (b) Velocity after 5 seconds (initial velocity = 0):
Using the impulse-momentum theorem:
\[
J = \Delta p = m \cdot v
\]
We can calculate the final velocity by finding the impulse and then solving for \( v \) (assuming initial velocity is 0).

#### (c) Velocity after 5 seconds (initial velocity = \(-2 \, \text{m/s}\)):
We use the same impulse-momentum theorem, but now the initial velocity is \(-2 \, \text{m/s}\), so we modify the equation for final velocity.

#### (d) Average Force from 0 to 5 seconds:
The average force is:
\[
F_{\text{avg}} = \frac{J}{\Delta t}
\]
where \( J \) is the impulse calculated earlier and \( \Delta t \) is the total time (5 seconds).

---

### **Bài 2:**
A 10 g bullet is fired at a 3 kg block of wood.

#### (a) Velocity of the bullet-wood system after collision:
Using the conservation of momentum:
\[
m_1 v_1 + m_2 v_2 = (m_1 + m_2) v
\]
where \( m_1 = 0.01 \, \text{kg} \) (bullet), \( v_1 = 240 \, \text{m/s} \), and \( m_2 = 3 \, \text{kg} \) (wood block, initially at rest).

#### (b) Movement after collision, considering friction:
The system slides on a surface with a coefficient of kinetic friction \( \mu = 0.25 \). We can find the distance the system moves before stopping using work-energy principles:
\[
W_{\text{friction}} = \Delta K
\]

---

### **Bài 3:**
The angular position of a wheel is given as \( \theta(t) = 5 + 10t + 2t^2 \). We need to calculate the angular velocity and acceleration at \( t = 0 \) and \( t = 3 \, \text{s} \).

#### (a) Angular velocity:
The angular velocity \( \omega(t) \) is the derivative of \( \theta(t) \) with respect to time:
\[
\omega(t) = \frac{d\theta}{dt}
\]

#### (b) Angular acceleration:
The angular acceleration \( \alpha(t) \) is the derivative of \( \omega(t) \):
\[
\alpha(t) = \frac{d\omega}{dt}
\]

---

Would you like me to go through the detailed calculations for any specific problem or part? Here are five related questions that might be useful:

1. How is the concept of impulse related to momentum?
2. What is the significance of the area under a force-time graph?
3. How can you calculate the velocity of a system after a collision using conservation of momentum?
4. How does the coefficient of friction affect the motion of an object on a surface?
5. What is the difference between angular velocity and angular acceleration?

**Tip:** When dealing with force-time graphs, always calculate the area under the curve for impulse, as this will help in finding changes in momentum.

Problems related to force, impulse, momentum, and angular motion involving a mass, a bullet and a wheel's angular position.

Solve problems related to impulse, momentum, and angular motion involving a mass of 2.5 kg, a bullet fired into a wooden block, and the angular motion of a wheel. Key concepts include the impulse-momentum theorem, conservation of momentum, and angular velocity/acceleration. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation