Question

part ii. problem solving (10 pts)
show your solution clearly.

1. a car travels 100 meters in 5 seconds. what is its speed?
2. an object moves 200 meters north in 20 seconds. what is its velocity?
3. a runner speeds up from 2 m/s to 8 m/s in 3 seconds. what is the acceleration?
4. find the force when a 10 kg object accelerates at 2 m/s².
5. a 20 n force pushes right while a 10 n force pushes left. what is the net force and direction?
6. a 2 kg ball and a 4 kg ball are dropped from the same height. which hits first (ignore air resistance)?
7. two billiard balls collide elastically. describe what happens to their kinetic energy.
8. a car crashes into another and they stick together. what type of collision is this?
9. what happens to the gravitational force if the distance between two objects doubles?
10. a 50 kg student is pulled by gravity (9.8 m/s²). calculate their weight in newtons.

Explanation:

Step1: Recall the formula for speed

Speed \(v=\frac{d}{t}\), where \(d = 100\) meters and \(t = 5\) seconds.

Step2: Substitute values into the formula

\(v=\frac{100}{5}\)

Step3: Calculate the result

\(v = 20\) m/s

Step1: Recall the formula for velocity

Velocity \(v=\frac{d}{t}\), where \(d = 200\) meters (direction: north) and \(t = 20\) seconds.

Step2: Substitute values into the formula

\(v=\frac{200}{20}\)

Step3: Calculate the result

\(v = 10\) m/s north

Step1: Recall the formula for acceleration

Acceleration \(a=\frac{v_f - v_i}{t}\), where \(v_f=8\) m/s, \(v_i = 2\) m/s, and \(t = 3\) seconds.

Step2: Substitute values into the formula

\(a=\frac{8 - 2}{3}\)

Step3: Calculate the result

\(a=\frac{6}{3}=2\) m/s²

Step1: Recall Newton's second law \(F = ma\)

Here \(m = 10\) kg and \(a=2\) m/s².

Step2: Substitute values into the formula

\(F=10\times2\)

Step3: Calculate the result

\(F = 20\) N

Step1: Determine net force for opposite - direction forces

If \(F_1 = 20\) N (right) and \(F_2=10\) N (left), net force \(F_{net}=F_1 - F_2\) (taking right as positive).

Step2: Substitute values into the formula

\(F_{net}=20 - 10\)

Step3: Calculate the result

\(F_{net}=10\) N to the right

Step1: Use the equation of motion for free - fall \(h = v_0t+\frac{1}{2}gt^2\) (since \(v_0 = 0\), \(h=\frac{1}{2}gt^2\), \(t=\sqrt{\frac{2h}{g}}\))

Mass does not appear in the formula for the time of free - fall (\(t=\sqrt{\frac{2h}{g}}\), where \(h\) is height and \(g\) is acceleration due to gravity).

Step2: Conclusion

Both balls hit the ground at the same time.

Step1: Recall the property of elastic collision

In an elastic collision, kinetic energy is conserved.

Step1: Recall the definition of in - elastic collision

When two objects stick together after a collision, it is a perfectly in - elastic collision.

Step1: Recall the formula for gravitational force \(F_g=\frac{Gm_1m_2}{r^2}\)

If \(r' = 2r\), then \(F_g'=\frac{Gm_1m_2}{(2r)^2}\)

Step2: Simplify the formula

\(F_g'=\frac{1}{4}\times\frac{Gm_1m_2}{r^2}\)

Step3: Conclusion

The gravitational force is reduced to \(\frac{1}{4}\) of its original value.

Step1: Recall the formula for weight \(W = mg\)

Here \(m = 50\) kg and \(g = 9.8\) m/s².

Step2: Substitute values into the formula

\(W=50\times9.8\)

Step3: Calculate the result

\(W = 490\) N

Answer:

1. \(20\) m/s
2. \(10\) m/s north
3. \(2\) m/s²
4. \(20\) N
5. \(10\) N to the right
6. Both balls hit the ground at the same time.
7. Kinetic energy is conserved.
8. Perfectly in - elastic collision
9. The gravitational force is reduced to \(\frac{1}{4}\) of its original value.
10. \(490\) N