Lee Majors, a professional stuntman, is preparing to run off a dock and dive into a large kayak. The stuntman has a mass of 80.8 kg and the kayak has a mass of 29.9 kg. If Lee dives horizontally with a velocity of 5.50 m/s, what is his velocity immediately after he jumps into the kayak?
Ignore velocities in the y direction.
What I did was (80.8)(v) + 29.9(5.50) = 0 and solved for v. Got -2.04, which was marked incorrect. Where am I going wrong and how do I correct it?
Thank you!
The kayak is initially stationary(V2 = 0). I'm assuming their velocities are equal after landing.
M1*V1 + M2*V2 = M1*V + M2*V.
80.8*5.5 + 29.9*0 = 80.8V + 29.9V.
444.4 = 110.7V,
V = 4.01 m/s.
Thank you!
To determine Lee Major's velocity immediately after he jumps into the kayak, you need to apply the law of conservation of momentum.
The law of conservation of momentum states that the total momentum before an event is equal to the total momentum after the event, assuming no external forces are acting on the system.
In this case, the momentum before the jump is given by the equation:
(mass of Lee Majors) * (initial velocity of Lee Majors) + (mass of the kayak) * (initial velocity of the kayak) = (mass of Lee Majors) * (final velocity of Lee Majors) + (mass of the kayak) * (final velocity of the kayak)
Let's plug in the given values:
(80.8 kg) * (5.50 m/s) + (29.9 kg) * (0 m/s) = (80.8 kg) * (final velocity of Lee Majors) + (29.9 kg) * (final velocity of the kayak)
Simplifying the equation:
444.4 kg·m/s = (80.8 kg) * (final velocity of Lee Majors) + 0 kg·m/s
To find the final velocity of Lee Majors, rearrange the equation:
(final velocity of Lee Majors) = (444.4 kg·m/s) / (80.8 kg)
Calculating:
(final velocity of Lee Majors) ≈ 5.49 m/s
So, the correct answer is approximately 5.49 m/s, not -2.04 m/s.
It seems you made an error in your calculation, resulting in a negative sign. Check your calculations and make sure you are using the correct values.