In a stunt, three people jump off a platform and fall 9.9 m onto a large air bag. A fourth person at the other end of the air bag, the "flier," is launched 19 m vertically into the air. Assume all four people have a mass of 67 kg.
a. What fraction of the mechanical energy of the three jumpers is transferred to the mechanical energy of the flier?
(Express your answer to two significant figures.)
b. What is the total momentum of the three jumpers just before they land on the air bag? Let upward be the positive direction.
(Express your answer to two significant figures and include appropriate units.)
c. What is the momentum of the flier just after launch? Let upward be the positive direction.
(Express your answer to two significant figures and include appropriate units.)
a. Initial gravitational potential energy = 3*67 kg*9.8 m/s^2*9.9 m = 19501 J
Final GPE = 1*67 kg*9.8 m/s^2*19 m = 12475 J
fraction of the mechanical energy = 12475 J / 19501 J = 64/100 = 0.64
b. 3 jumper,s initial gravitational potential energy = final kinetic energy
19501 J = 1/2 * 3*67 kg*v^2
v^2 = 2*19501 J / (3*67 kg) = 194 J/kg
v = -13.93 m/s
Initial momentum = 3*67 kg*(-13.93 m/s) = -2800 kg.m/s
c. flier,s final GPE = initial KE
12475 J = 1/2 *67 kg*v^2
v^2 = 2*12475 J / 67 kg = 373.4 J/kg
v = 19.30 m/s
Initial momentum = 67 kg*19.30 m/s = 1293 kg.m/s ~= 1300 kg.m/s
To solve this problem, we can use the principles of conservation of energy and conservation of momentum.
a. To find the fraction of mechanical energy transferred to the flier, we need to calculate the mechanical energies before and after the jump.
The mechanical energy before the jump is the potential energy at the initial height:
E_initial = m * g * h
where m is the mass of the jumpers (67 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (9.9 m).
The mechanical energy after the jump is the sum of the potential energy at the final height and the kinetic energy of the flier:
E_final = m * g * H + 1/2 * m * v^2
where H is the height reached by the flier (19 m) and v is the velocity of the flier at that height.
The fraction of mechanical energy transferred to the flier is given by:
Fraction = (E_final - E_initial) / E_initial
Plugging in the given values and solving, we get:
Fraction = ((67 kg * 9.8 m/s^2 * 19 m) + (1/2 * 67 kg * v^2) - (67 kg * 9.8 m/s^2 * 9.9 m)) / (67 kg * 9.8 m/s^2 * 9.9 m)
b. To find the total momentum of the three jumpers just before they land on the air bag, we need to find the momentum of each jumper and add them up.
The momentum of each jumper is given by:
p = m * v
where m is the mass of the jumper (67 kg) and v is their velocity just before landing.
The total momentum is the sum of the individual momenta.
c. To find the momentum of the flier just after launch, we can use the principle of conservation of momentum. Since no external horizontal forces are acting on the system, the total momentum of the system should be conserved.
The momentum of the flier just after launch is given by:
p_flier = - (p_initial_1 + p_initial_2 + p_initial_3)
where p_initial_1, p_initial_2, and p_initial_3 are the initial momenta of the three jumpers.
Let's calculate these values step by step.
To solve this problem, we need to apply the principles of energy and momentum conservation.
a. The fraction of mechanical energy transferred can be determined by comparing the initial mechanical energy of the three jumpers with the final mechanical energy of the flier.
The initial mechanical energy of the three jumpers is given by the potential energy formula: PE = mgh.
PE_initial = 3 * (67 kg) * (9.8 m/s^2) * (9.9 m) = 19356 J (rounded to four significant figures).
The final mechanical energy of the flier is also given by the potential energy formula: PE = mgh.
PE_final = (67 kg) * (9.8 m/s^2) * (19 m) = 12334 J (rounded to four significant figures).
The fraction of energy transferred is then: (PE_final / PE_initial) * 100%.
Fraction of energy transferred = (12334 J / 19356 J) * 100% = 63.7% (rounded to two significant figures).
Therefore, approximately 63.7% of the mechanical energy of the three jumpers is transferred to the mechanical energy of the flier.
b. The total momentum of the three jumpers just before they land on the air bag can be calculated by summing their individual momenta.
Momentum (p) = mass (m) * velocity (v).
The initial velocity of the three jumpers is assumed to be zero since they are at rest before jumping. Therefore, their momenta just before landing are all zero.
Total momentum of the three jumpers = 0 kg*m/s (rounded to two significant figures).
c. The momentum of the flier just after launch can be calculated using the momentum principle.
The final velocity of the flier, just after launch, can be determined using the formula: v = √(2gh), where h is the vertical distance traveled.
Final velocity (v) = √(2 * 9.8 m/s^2 * 19 m) = 19 m/s (rounded to two significant figures).
Therefore, the momentum of the flier just after launch is:
Momentum (p) = mass (m) * velocity (v).
Momentum of the flier = (67 kg) * (19 m/s) = 1273 kg*m/s (rounded to two significant figures).
Therefore, the momentum of the flier just after launch is approximately 1273 kg*m/s.