An offensive tackle (Bob) has a mass is 137.8 kg and is running at a velocity of
ID: 1322151 • Letter: A
Question
An offensive tackle (Bob) has a mass is 137.8 kg and is running at a velocity of 4.28 m/s toward Joe. A defensive guard (Joe) is 123.65 kg and is running at 4.92 m/s toward and collinear with Bob. They will eventually have an inelastic collision.
How much force is required for Bob (offensive tackle) to achieve and acceleration of 15.87 m/s^2? (1 pt)
What is the momentum of Bob, the offensive tackle, before the collision?
What is the momentum of Joe, the defensive guard, before the collision?
What is the total momentum of the system (Bob and Joe) before the collision?
Bob wrap tackles Joe, which results in an inelastic collision. What is the velocity of the system (Bob and Joe) after the collision?
What is the momentum of the system (Bob and Joe) after the collision?
The answers in questions 6 & 7 are explained by the principle of what????
Explanation / Answer
a) F=ma; F=137.8*15.87
F= 20187N
b) m=mv=137.8*4.28
m= 589.8 kg.m/s
c) m=mv=123.65*-4.92
? m= -608.4 kg.m/s
d) mt=589.8 - 608.4
mt= -18.6 kg.m/s
e) v=(m1*u1+m2*u2)/(m1+m2)
v=(137.8*4.28+123.65*-4.92)/(137.8+123.65)
v=-0.07 m/s
f) m=261.45*-0.07
m=-18.6 kg.m/s
g) conservation of momentum