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 if kinetic energy is conserved. In an elastic collision, kinetic energy is conserved; in an inelastic collision, it is not. First, calculate initial and final kinetic energy for a trial (e.g., Trial 1). Initial KE of impact car: \( KE_i=\frac{1}{2}m_iv_i^2=\frac{1}{2}(1000)(20)^2 = 200000\space J\). Final KE: \( KE_f=\frac{1}{2}(1000)(5)^2+\frac{1}{2}m_{Tesla}(7.5)^2\). Since Tesla was at rest, initial KE of Tesla is 0. As final KE is less than initial (energy lost, e.g., heat, deformation), it's an inelastic collision (also, cars don't stick, so it's a partially inelastic collision, but generally called inelastic as KE isn't conserved).

Step1: Recall momentum formula

Momentum \( p = mv \). The momentum gained by Tesla is equal to the momentum lost by the impact car (conservation of momentum, since system momentum is conserved). Initial momentum of impact car: \( p_{i - impact}=m_iv_i=(1000)(20)=20000\space kg\cdot m/s\). Final momentum of impact car: \( p_{f - impact}=m_iv_{f - impact}=(1000)(5) = 5000\space kg\cdot m/s\).

Step2: Calculate momentum lost (gained by Tesla)

Momentum gained by Tesla \( \Delta p = p_{i - impact}-p_{f - impact}=20000 - 5000 = 15000\space kg\cdot m/s\). (Alternatively, \( p_{Tesla}=m_{Tesla}v_{Tesla}\), but since we don't know \( m_{Tesla}\), using conservation is better. Wait, but from the data, in all trials, Tesla's velocity after collision is 7.5 m/s. Wait, maybe we can also calculate \( p_{Tesla}=m_{Tesla}(7.5)\), and since \( \Delta p_{impact}=-\Delta p_{Tesla}\), so \( \Delta p_{Tesla}=m_i(v_i - v_{f - impact})\). For Trial 1: \( 1000(20 - 5)=1000\times15 = 15000\space kg\cdot m/s\).

Step1: Use momentum change formula

Momentum gained by Tesla = momentum lost by impact car. Formula: \( \Delta p = m_i(v_i - v_{f - impact})\). For Trial 2, \( m_i = 1500\space kg\), \( v_i = 20\space m/s\), \( v_{f - impact}=10\space m/s\).

Step2: Substitute values

\( \Delta p=1500(20 - 10)=1500\times10 = 15000\space kg\cdot m/s\).

Answer:

Inelastic Collision

Part b i