Impulse and Dodgeball

By: Sravya Madabhushi

G-block Physics

What is Impulse? What is Impulse? In physics, the In physics, the

quantity quantity Force• timeForce• time is known is known

as as impulseimpulse. And . And since the quantity since the quantity m•v is the m•v is the momentum, the momentum, the quantity m• v must quantity m• v must be the be the change in change in momentummomentum. .

Therefore Therefore Impulse = Change in Impulse = Change in

momenmomenttumum

Collisions in Dodgeball

The types of collision involved in the sport is an inelastic and elastic collision. For instance when two balls collide and bounce away form each other, or when a player is hit by a ball.

In an perfectly elastic collision two objects return to their original shape and move away from the collision separately. An elastic collision is a collision where total momentum and total kinetic energy are both conserved.

In a perfectly inelastic collision, two objects stick to together and move as one mass after the collision. Momentum is conserved but kinetic energy is not. In an inelastic collision, kinetic energy, which is not conserved, is converted into internal inelastic potential energy (when objects deform). The remainder of the KE is converted into heat and sound energy.

Equation 1Equation 1ΔΔP= mVP= mVff-mV-mVii

Impulse is the change in Impulse is the change in linear momentum. linear momentum.

P on the left hand side of P on the left hand side of the equation stands for the equation stands for impulse which is given in impulse which is given in Newton-seconds or Newton-seconds or kg*m/skg*m/s

On the right hand side is On the right hand side is mass times final velocity mass times final velocity minus mass times initial minus mass times initial velocity. velocity.

Example 1Example 1According to the official rules of According to the official rules of Dodgeball, 6 regulation size balls Dodgeball, 6 regulation size balls should be used: 4 Blockers™ should be used: 4 Blockers™ (8.5") and 2 Stingers™ (5"). The (8.5") and 2 Stingers™ (5"). The larger balls have a mass of .65 kg. larger balls have a mass of .65 kg. Suppose during a game, a player Suppose during a game, a player throws a dodgeball at 15 m/s, throws a dodgeball at 15 m/s, which hits another stationary ball which hits another stationary ball on the ground. The first ball on the ground. The first ball continues to travel in the same continues to travel in the same direction after the collision but at a direction after the collision but at a slower velocity of 10 m/s. slower velocity of 10 m/s. Therefore, the change in Therefore, the change in momentum or impulse on the first momentum or impulse on the first ball is P= (.65)(10)-(.65)(15) = -ball is P= (.65)(10)-(.65)(15) = -3.25 kg*m/s.3.25 kg*m/s.This collision is still an inelastic This collision is still an inelastic collision, however the second collision, however the second ball’s velocities were not taken ball’s velocities were not taken into account, for the purpose of into account, for the purpose of simplifying the problem.simplifying the problem.

Example 1 continuedExample 1 continued

Use conservation of momentum to calculate the velocity of Use conservation of momentum to calculate the velocity of the second ball after the collision:the second ball after the collision:m_1v_1i + m_2v_2i = m_1v_1f + m_2v_2fm_1v_1i + m_2v_2i = m_1v_1f + m_2v_2fBecause mass is the same for both the balls, we can cancel Because mass is the same for both the balls, we can cancel mass from both sides of the equation.mass from both sides of the equation.v_1i = 15 m/s, v_2i = 0 m/s , v_1f = 10 m/s , v_2f = ???v_1i = 15 m/s, v_2i = 0 m/s , v_1f = 10 m/s , v_2f = ???V_2f= v_1i + v_2i - v_1f = (15) + (0) – (10) = 5 m/sV_2f= v_1i + v_2i - v_1f = (15) + (0) – (10) = 5 m/sTherefore V2final is 5 m/s and momentum is conserved.Therefore V2final is 5 m/s and momentum is conserved.Delta KE= KEf- KEiDelta KE= KEf- KEiKEi= .5m_1v_1i^2+.5m_2v_2i^2 KEi= .5m_1v_1i^2+.5m_2v_2i^2 KEi= 73.125 JKEi= 73.125 JKEf= .5(.65)(10)^2+.5(.65)(5)^2= 40.625 JKEf= .5(.65)(10)^2+.5(.65)(5)^2= 40.625 JDelta KE= 40.625 J- 73.125J= -32.5J =Decrease in KEDelta KE= 40.625 J- 73.125J= -32.5J =Decrease in KETherefore the collision is inelasticTherefore the collision is inelastic

Impulse Equation #2

Impulse= Force x change in time.

F is the constant total net force applied,

t is the time interval over which the force is applied.

P is the change in momentum of an object to which a force is applied

Example 2

If smart player was inevitably to get bombarded by a ball, he would move in the same direction as the ball that is heading towards him. This would increase in the time interval between the release of the ball and its contact with a body, therefore reducing its force. This would minimize injury to the player.

Remember that Force is inversely proportional to time and impulse.

http://www.youtube.com/watch?v=dhKQ-whmSBA

Example 2 Continued For example, A ball with a mass of .65 kg is headed towards a

player with an impulse of 6 kg*m/s. The player moves along in the same direction of the ball manages to avoid a collision for 3 sec. What is the force or the ball when it collides with the player?

ΔP= F ΔT. F= ΔP/ ΔT F=6 kg*m/s / F(3) F= 3 Netwons

Ideally, this collision is elastic, because when the ball come in contact with the player it will bounce off and travel in the opposite direction. The player having received an impact from the ball will move backwards slightly as a result. It is important to note that very few collisions in real life are perfectly elastic or perfectly inelastic; most collision fall somewhere in the middle.

Bibliography• The National Dodgeball League

http://www.thendl.com/• http://www.physicsclassroom.com/

class/momentum/U4l1b.cfm• http://www.euclideanspace.com/

physics/dynamics/collision/impulse/index.htm