A popular school-yard brawl maneuver is the "one-two finish," a maneuver known to brutally knockout the opponent. It consists of an uppercut punch, and, when the opponent's body is drifting through the air, the attacker prepares a roundhouse kick to finish his opponent. A particular brawl breaks out at Genghis Khan Academy and a student decides to attempt the maneuver on a mean-spirited bully. The student needs 0.5 seconds to prepare his roundhouse kick. At what velocity must the 45 kg boy punch to keep the 80 kg bully in the air long enough to ready his roundhouse kick? Assume the momentum transfer is elastic and the fist's velocity after punching is 0.5 m/sec. If the punch takes place in 0.25 seconds, what force must the student exert?
First, we must consider the bully's motion through the air. After the bully is punched, there are no forces acting upon him besides gravity, so he is in freefall. He will rise initially, and then he will descend back to the ground. This motion can be expressed in a parabolic position-time graph. The graph, pictured below, rises initially then returns to the axis, showing the bully's take off and return to level ground.
We know that this motion is symmetric (property of a parabola). This means the initial velocity equals the final velocity. The equation that describes motion of an object in freefall is the equation y= (1/2)gt^2 + Vot. We know that the position is 0 at the start (when the boy punched him) and at the end (when the bully lands back on the ground). If we want the bully to land at 0.5 seconds so the roundhouse kick can be delivered, we can substitute it for t.
0 = (1/2) (-9.8) (0.5)^2 + Vo(0.5)
1.225 = (0.5) Vo
Vo = 2.45 m/sec
1.225 = (0.5) Vo
Vo = 2.45 m/sec
So, the bully must leave the ground at 2.45 m/sec. Now, we can consider the conservation of momentum during the punch. To do this, we can construct a momentum bar chart to qualitatively see the transfer of momentum.
We can write a momentum conservation equation and then solve for the necessary initial velocity. The equation, seen below, means that some of the Boy's initial momentum of the punch was transferred to the bully and some was kept in the boy's fist, but all of the momentum was preserved.
Pboy + Pbully = Pboy' + Pbully'
(45 kg) ( x m/sec) + (80 kg) (0 m/sec) = (45 kg) (0.5 m/sec) + (80 kg) (2.45 m/sec)
45x kg*m/sec = 218.5 kg*m/sec
x = 4.856 m/sec
(45 kg) ( x m/sec) + (80 kg) (0 m/sec) = (45 kg) (0.5 m/sec) + (80 kg) (2.45 m/sec)
45x kg*m/sec = 218.5 kg*m/sec
x = 4.856 m/sec
So, the boy's fist must make contact with the bully at 4.856 m/sec in order for the boy to have enough time to perform his roundhouse kick.
Now, we can consider the impulse needed for this action to take place. We know that the bully's velocity changes from 0 to 2.45 m/sec. We can use this to calculate Δv, or change in velocity. Δv = Vfinal - Vinitial = 2.45 m/sec - 0 m/sec = 2.45 m/sec.
We are also given the time over which the velocity change occurs and the bully's mass. We can plug these values into the impulse-momentum equation to solve for the final variable, force.
Now, we can consider the impulse needed for this action to take place. We know that the bully's velocity changes from 0 to 2.45 m/sec. We can use this to calculate Δv, or change in velocity. Δv = Vfinal - Vinitial = 2.45 m/sec - 0 m/sec = 2.45 m/sec.
We are also given the time over which the velocity change occurs and the bully's mass. We can plug these values into the impulse-momentum equation to solve for the final variable, force.
F * Δt = m* Δv
F * (0.25 sec) = 80 kg (2.45 m/sec)
F = 784N
A force of 784N is needed for the boy to punch the bully and change his velocity enough for him to rise up for 0.5 seconds and then be taken out by a roundhouse kick.
F * (0.25 sec) = 80 kg (2.45 m/sec)
F = 784N
A force of 784N is needed for the boy to punch the bully and change his velocity enough for him to rise up for 0.5 seconds and then be taken out by a roundhouse kick.