I need help with working out the answers to these questions: First A 930 k g spo
ID: 2262893 • Letter: I
Question
I need help with working out the answers to these questions: First
A 930kg sports car collides into the rear end of a 2400kg SUV stopped at a red light. The bumpers lock, the brakes are locked, and the two cars skid forward 3.0m before stopping. The police officer, knowing that the coefficient of kinetic friction between tires and road is 0.80, calculates the speed of the sports car at impact. What was that speed?
Then I also need to know
In a ballistic pendulum experiment, projectile 1 results in a maximum height h of the pendulum equal to 2.2cm . A second projectile causes the pendulum to swing twice as high, h2 = 4.4cm . The second projectile was how many times faster than the first?
Explanation / Answer
KE after impact = work done by friction to stop the vehicles
1/2*(m+M) V^2 = u * (M+m)*g*d
V: Speed of locked together vehicles just after impact
d: skid distance =3.0 m
m,M masses of sports car and SUV respectively. = 930 kg, 2400 Kg
u = 0.80
Solving for V
1/2*(930+2400) V^2 = 0.80 * (2400+930)*9.81*3
v^2=47.088
V=6.86 m/s
Momentum equation for impact is
m*v = (m+M)*V
v: Speed of sports car before impact
v={(m+M)*V }/m
v= 22843.8/930
v=24.56 m/s
part 2)
The height is proportional to the kinetic energy of the projectile. So the ratio of the two KEs is 4.4/2.2 = 2
KE is proportional to velocity squared, so velocity is proportional to the square root of the KE, which is sqrt(2) = 1.414, which is the answer.
In more detail
KE = PE = mgh, which shows you the KE and height proportionality.
KE = 0.5mV^2 or
V = sqrt(2KE/m) which shows you the other proportionality
combining
V = sqrt(2mgh/m) = sqrt(2gh)
V1/V2 = sqrt(2gh1) / sqrt(2gh2)
V1/V2= sqrt(h1/h2)
= sqrt(4.4/2.2) = 1.414