Collisons, energy?

Question

1) A 1.0X10^3kg Toyota collides into the rear end of a 2.2X10^3kg Cadillac stopped at a red light. The bumpers lock, the brakes are locked, and the cars skid forward 2.8m before stopping. The police officer, knowing that the coefficient of friction between the tires and the road is 0.40, calculates the speed of the Toyota at impact. What is it?

2) A 975kg rocket is traveling at a speed of 5.8X10^3m/s with respect top the Earth when a pre-designed explosion separates the rocket into 2 sections of equal mass that then move with a relative speed (relative to each other) of 2.2X10^3m/s along the original line of motion.

(a) What is the speed and direction of each section (relative to Earth) after the explosion?

(b) How much energy was supplied by the explosion? [Hint: What is the change in Ek as a result of the explosion?]

3) What is the force exerted on a rocket, given that the propelling gases are expelled at a rate of 13000kg/s with a speed of 40000m/s(at the moment of takeoff)?

Answer 1

You need to use the conservation of momentum at all three problems.

1. If the speed of the Toyota is 'v1', and of the wreck right after impact is 'v':

m1 * v1 = (m1 + m2) * v

where m1 = 1000 kg, m2 = 2200 kg

To calculate 'v' we use the law of conservation of energy:

(m1+m2) * v^2 / 2 = u * (m1+m2) * g * d

where u = 0.40 the friction coefficient, d = 2.8 m the stopping distance, g = 9.8 m/s2. Calculate v and plug in the first equation to obtain v1.

2. m/2 * v1 + m/2 * v2 = m * v

where m = 975 kg, v = 5800 m/s and v1-v2 = 2200 m/s

a. To obtain the speeds relative to Earth, calculate v1 and v2

b. Use the conservation of energy. The supplied energy:

E = m * v^2 / 2 - m/2 * v1^2 / 2 - m/2 * v2^2 / 2

3. The change of momentum per unit time is dp = dm * dv, so the rate of momentum change (= force) is:

F = dm * dv / dt

dm = 13000 kg/s, dv = 40000 m/s, dt = 1, you obtain the result in kg*m/s2, and remember that 1 kg*m/s2 = 1 N.