A Car Of 400 Kg Mass Traveling At 20 M/S Collides With
1 A Car A Mass Of 400 Kg Travels At 20 Ms Teh Car Collides With A S
A car with a mass of 400 kg is traveling at 20 m/s and collides with a stationary truck of mass 1600 kg. The vehicles interlock upon collision.
Assignment Instructions:
1. Calculate the velocity of the combined car and truck system after the collision using appropriate equations.
2. For each calculation, show the equation used, substitute the given numbers, simplify, and solve.
3. Provide the final velocity in appropriate units.
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Paper For Above instruction
Introduction
Understanding the principles of momentum and collisions is central to physics, especially in the context of vehicle accidents and safety engineering. The problem involves applying the law of conservation of momentum to determine the velocity of interconnected vehicles post-collision. Similarly, the subsequent scenarios involve momentum conservation to compute recoil velocities, momentum before and after impacts, and the effect of forces over time. This paper details the calculations and physical principles involved in these problems.
1. Collision Between a Moving Car and a Stationary Truck
Given data:
- Mass of car, \(m_c = 400\,kg\)
- Velocity of car, \(v_c = 20\,m/s\)
- Mass of truck, \(m_t = 1600\,kg\)
- Initial velocity of truck, \(v_t = 0\,m/s\)
Objective:
Calculate the velocity of the combined system after collision, \(v_f\).
Method:
Using the law of conservation of momentum:
\[
m_c v_c + m_t v_t = (m_c + m_t) v_f
\]
since the truck is initially stationary, \(v_t=0\), simplifying to:
\[
m_c v_c = (m_c + m_t) v_f
\]
Rearranged:
\[
v_f = \frac{m_c v_c}{m_c + m_t}
\]
Calculation:
Plugging in the numbers:
\[
v_f = \frac{400\,kg \times 20\,m/s}{400\,kg + 1600\,kg} = \frac{8000}{2000} = 4\,m/s
\]
Result:
The velocity of the car-truck system after the collision is 4 m/s in the direction of the initial velocity of the car.
2. Recoil Velocity of a Launcher Due to Projectile Launch
Given data:
- Mass of projectile, \(m_p = 50\,kg\)
- Velocity of projectile, \(v_p = 800\,m/s\)
- Mass of launcher, \(m_l = 2000\,kg\)
- Initial velocity of launcher, \(v_{l,i} = 0\,m/s\)
Objective:
Determine the recoil velocity of the launcher after firing.
Method:
Applying conservation of momentum:
\[
\text{Initial total momentum} = \text{Final total momentum}
\]
\[
0 = m_p v_p + m_l v_{l,f}
\]
The launcher recoils in the opposite direction to the projectile, so:
\[
m_l v_{l,f} = - m_p v_p
\]
Rearranged:
\[
v_{l,f} = - \frac{m_p v_p}{m_l}
\]
Calculation:
Substituting values:
\[
v_{l,f} = - \frac{50\,kg \times 800\,m/s}{2000\,kg} = - \frac{40000}{2000} = -20\,m/s
\]
Result:
The recoil velocity of the launcher is 20 m/s in the opposite direction of the projectile.
3. Momentum and Velocity of Dog and Skateboard
Given data:
- Mass of dog, \(m_d = 24\,kg\)
- Speed of dog, \(v_d = 3\,m/s\)
- Mass of skateboard, \(m_s = 3.6\,kg\)
- Initial velocity of skateboard, \(v_s = 0\,m/s\)
a. Momentum of the dog before jumping onto the skateboard:
\[
p_d = m_d v_d = 24\,kg \times 3\,m/s = 72\,kg \cdot m/s
\]
b. Momentum of the dog and skateboard after jumping together:
Since momentum is conserved and the skateboard is initially stationary:
\[
p_{total} = p_d + p_s = 72 + 0 = 72\,kg \cdot m/s
\]
c. Velocity of the dog and skateboard together:
\[
v_{f} = \frac{p_{total}}{m_d + m_s} = \frac{72}{24 + 3.6} = \frac{72}{27.6} \approx 2.61\,m/s
\]
d. Momentum of the dog before jumping:
Already calculated as 72 kg·m/s.
e. Time taken to stop the combined system with an average force:
Using impulse-momentum theorem:
\[
F \times \Delta t = \Delta p
\]
where \(\Delta p = p_{final} - p_{initial} = 0 - 72 = -72\,kg \cdot m/s\)
Rearranged for \(\Delta t\):
\[
\Delta t = \frac{\Delta p}{F} = \frac{72\,kg \cdot m/s}{9\,N} = 8\,s
\]
Result:
It would take approximately 8 seconds for an average force of 9 N to bring the dog and skateboard to a stop.
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Conclusion
The calculations demonstrate the application of fundamental physics principles—conservation of momentum, impulse, and Newton’s laws—to real-world scenarios involving vehicles, projectiles, and objects in motion. The combined velocity of the car and truck post-collision is 4 m/s, and the launcher recoils at 20 m/s after firing the projectile. The dog and skateboard system moves at approximately 2.61 m/s after impact, and it takes about 8 seconds to halt this combined motion under a consistent force. These principles are essential for simplifying complex interactions and designing safety mechanisms, such as crumple zones in vehicles and recoil shielding in weaponry.
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