A. A train goes twice as fast downhill as it can go uphill, Ana ( 2)/(3) as fast uphill as it can go on level ground. If it goes 120 miles per hour downhill, how long will it take to travel 45 miles on flat land?
B. A 0.010 kg bullet traveling at 300.0 m/s east hits a motionless 2.0 kg block and bounces off it, retracing its original path with a velocity of 100.0 m/s west. What is the final velocity of the block? Assume there is no friction or air resistance.
The Correct Answer and Explanation is:
Sure! Let’s carefully solve both problems step-by-step with detailed explanations.
A. Train speed and time calculation
Problem restatement:
- Downhill speed = 120 mph
- Train goes twice as fast downhill as uphill
- Train goes 2/3 as fast uphill as on level ground
- Find how long it takes to travel 45 miles on flat ground
Step 1: Express uphill speed
Let the uphill speed be $V_u$.
Given downhill speed $V_d = 2 \times V_u$.
We know downhill speed $V_d = 120$ mph.
So,
$$
V_d = 2V_u \implies 120 = 2V_u \implies V_u = 60 \text{ mph}
$$
Step 2: Express flat speed
Train goes $\frac{2}{3}$ as fast uphill as on level ground:
$$
V_u = \frac{2}{3} V_f \implies 60 = \frac{2}{3} V_f
$$
Multiply both sides by $\frac{3}{2}$:
$$
V_f = 60 \times \frac{3}{2} = 90 \text{ mph}
$$
Step 3: Calculate time on flat ground
Distance $d = 45$ miles, speed on flat $V_f = 90$ mph.
Time $t = \frac{d}{V} = \frac{45}{90} = 0.5$ hours.
Answer A:
It takes 0.5 hours (or 30 minutes) to travel 45 miles on flat land.
B. Bullet and block collision (momentum conservation)
Problem restatement:
- Bullet mass $m_b = 0.010 \, kg$
- Bullet initial velocity $v_{b,i} = 300.0 \, m/s$ east
- Block mass $m_{bl} = 2.0 \, kg$
- Block initial velocity $v_{bl,i} = 0 \, m/s$
- Bullet bounces off, final velocity $v_{b,f} = 100.0 \, m/s$ west (opposite direction)
- Find block’s final velocity $v_{bl,f}$
Step 1: Define directions
- East is positive, west is negative.
- Bullet initial velocity $v_{b,i} = +300.0 \, m/s$
- Bullet final velocity $v_{b,f} = -100.0 \, m/s$ (since west)
Step 2: Use conservation of momentum
Total momentum before = total momentum after
$$
m_b v_{b,i} + m_{bl} v_{bl,i} = m_b v_{b,f} + m_{bl} v_{bl,f}
$$
Plug values:
$$
(0.010)(300.0) + (2.0)(0) = (0.010)(-100.0) + 2.0 \times v_{bl,f}
$$
Calculate:
$$
3.0 = -1.0 + 2.0 v_{bl,f}
$$
Add 1.0 to both sides:
$$
3.0 + 1.0 = 2.0 v_{bl,f} \implies 4.0 = 2.0 v_{bl,f}
$$
Solve for $v_{bl,f}$:
$$
v_{bl,f} = \frac{4.0}{2.0} = 2.0 \, m/s
$$
Final Answer B:
The block’s final velocity is 2.0 m/s east.
Explanation
Problem A Explanation:
This problem involves understanding relative speeds and how different conditions (uphill, downhill, flat) relate to each other. The key is setting up relationships:
- The downhill speed is twice the uphill speed.
- The uphill speed is two-thirds of the flat speed.
By using the given downhill speed of 120 mph, you can find the uphill speed by dividing by two (since downhill is twice uphill). Then, since uphill speed is 2/3 of flat speed, use the uphill speed to calculate flat speed. With flat speed known, use the formula for time (distance/speed) to calculate how long it will take to travel 45 miles on flat ground.
The calculation shows that the flat speed is 90 mph, so traveling 45 miles takes 0.5 hours (or 30 minutes). This makes intuitive sense because at 90 mph, you cover 90 miles in one hour, so half that distance (45 miles) would take half an hour.
Problem B Explanation:
This problem involves an elastic collision where a bullet hits a block and bounces back, reversing direction. Because the bullet bounces, its final velocity is negative relative to the initial positive direction.
The law of conservation of momentum states that in a closed system with no external forces, the total momentum before and after the collision remains constant.
Momentum is the product of mass and velocity. The bullet initially moves east with positive velocity, and the block is stationary. After the collision, the bullet moves west (negative velocity) and the block moves east (unknown velocity).
By setting the total initial momentum equal to total final momentum and plugging in the known masses and velocities, you can solve for the block’s final velocity. The calculation shows the block moves east at 2.0 m/s.
This result makes physical sense: the bullet’s backward bounce imparts forward momentum to the block. Despite the bullet’s small mass, its high velocity transfers enough momentum to move the much heavier block at a measurable speed.