Question

1. engineers crash testing cars gathered the following data by running cars of different sizes into the newest tesla design. the tesla was at rest before the collision, and the impact car was moving at twenty meters per second.

trial	mass of impact car (kg)	velocity of impact car after collision (m/s)	velocity of tesla after collision (m/s)
2	1500	10	7.5
3	2000	12.5	7.5
4	2500	14	7.5

a. what type of collision occurs between the impact car and the tesla?

b. in each of these situations, the tesla gained momentum from the collision.
i. calculate the momentum gained from trial one.
ii. calculate the momentum gained from trial two.
iii. calculate the momentum gained from trial three.

Explanation:

Part a

To determine the collision type, we check momentum conservation (since collisions conserve momentum) and kinetic energy. For an elastic collision, kinetic energy is conserved; for inelastic, it's not. First, find the mass of Tesla (let \( m_T \)) and use momentum conservation \( m_I v_{I,i} = m_I v_{I,f} + m_T v_{T,f} \). From trial 1: \( 1000 \times 20 = 1000 \times 5 + m_T \times 7.5 \), solving gives \( m_T = 2000 \, \text{kg} \). Now check kinetic energy (KE) before and after. Initial KE: \( \frac{1}{2} \times 1000 \times 20^2 = 200000 \, \text{J} \). Final KE: \( \frac{1}{2} \times 1000 \times 5^2 + \frac{1}{2} \times 2000 \times 7.5^2 = 12500 + 56250 = 68750 \, \text{J} \). KE is not conserved, so it's an inelastic collision (since objects don't stick, it's a partially inelastic collision, but generally inelastic as KE is lost).

Step1: Recall Momentum Formula

Momentum \( p = mv \). The Tesla was at rest initially, so momentum gained is \( p = m_T v_{T,f} \). First, find \( m_T \) using momentum conservation for trial 1: \( m_I v_{I,i} = m_I v_{I,f} + m_T v_{T,f} \).
\( 1000 \times 20 = 1000 \times 5 + m_T \times 7.5 \)
\( 20000 = 5000 + 7.5 m_T \)
\( 7.5 m_T = 15000 \)
\( m_T = 2000 \, \text{kg} \).

Step2: Calculate Tesla's Momentum

Now, momentum gained by Tesla in trial 1 is \( p = m_T v_{T,f} = 2000 \times 7.5 = 15000 \, \text{kg·m/s} \). (Alternatively, use the change in momentum of the impact car: \( \Delta p_I = m_I (v_{I,i} - v_{I,f}) = 1000(20 - 5) = 15000 \, \text{kg·m/s} \), and by conservation, this equals Tesla's gained momentum.)

Step1: Use Momentum Conservation (or Impact Car's Momentum Change)

Momentum gained by Tesla equals the momentum lost by the impact car. For trial 2, \( \Delta p_I = m_I (v_{I,i} - v_{I,f}) \).
\( m_I = 1500 \, \text{kg} \), \( v_{I,i} = 20 \, \text{m/s} \), \( v_{I,f} = 10 \, \text{m/s} \).

Step2: Calculate Momentum Change

\( \Delta p_I = 1500(20 - 10) = 1500 \times 10 = 15000 \, \text{kg·m/s} \). (Or use \( p = m_T v_{T,f} \), \( m_T = 2000 \, \text{kg} \), \( v_{T,f} = 7.5 \, \text{m/s} \), so \( p = 2000 \times 7.5 = 15000 \, \text{kg·m/s} \).)

Answer:

Inelastic Collision (specifically, a partially inelastic collision as the cars do not stick together but kinetic energy is not conserved)

Part b i