Momentum conservation and transfer
Problem#1:
A 160.0 gram tennis ball is hurled towards a wall at 50 m/s. The ball bounces directly to the thrower at 60 m/s. What change in momentum does the ball undergo? Since the tennis ball contacts the well for 6.0*10^-3 seconds, find the average force exerted on the ball during the collision.
Before and After Momentum Equations:
P1 + P2 = P1 + P2
x + (.160 kg)*(50 m/s) = (.160 kg)*(-60 m/s)
x + 8 kgm/s = -9.8 kgm/s
x = -17.6 kgm/s
(which would be 17.6 kgm/s towards the thrower)
The only way to determine the change in momentum of a system is to use the momentum conservation equation. Since momentum is indicated by the letter P, and momentum is the mass of an object multiplied by the velocity , all the numbers plug in correctly. All is left then, is to simplify. This entire equation solves to be 17.6 kgm/s which is the speed that the tennis ball is traveling back towards the thrower.
P1 + P2 = P1 + P2
x + (.160 kg)*(50 m/s) = (.160 kg)*(-60 m/s)
x + 8 kgm/s = -9.8 kgm/s
x = -17.6 kgm/s
(which would be 17.6 kgm/s towards the thrower)
The only way to determine the change in momentum of a system is to use the momentum conservation equation. Since momentum is indicated by the letter P, and momentum is the mass of an object multiplied by the velocity , all the numbers plug in correctly. All is left then, is to simplify. This entire equation solves to be 17.6 kgm/s which is the speed that the tennis ball is traveling back towards the thrower.
The Average Exerted Force: ΣF = m*a
F (6.0*10^-3 s) = (.160 kg)(-60 m/s—50 m/s)
F (6.0*10^-3 s) = -17.6 kgm/s
F = -2933.33 N
To find the average force exerted upon the tennis ball, we use the equation above (ΣF = m*a), and then plug in all the values mentioned earlier in the problem, for mass and velocity. ΣF can be simplified to: F multiplied by the time that the tennis ball was touching the wall (6.0*10^-3 seconds). Once all the numbers are plugged in, they are simplified down to find that the average force exerted upon the tennis ball is -2933.33 N (the negative sign accounts for the direction of the ball).
F (6.0*10^-3 s) = (.160 kg)(-60 m/s—50 m/s)
F (6.0*10^-3 s) = -17.6 kgm/s
F = -2933.33 N
To find the average force exerted upon the tennis ball, we use the equation above (ΣF = m*a), and then plug in all the values mentioned earlier in the problem, for mass and velocity. ΣF can be simplified to: F multiplied by the time that the tennis ball was touching the wall (6.0*10^-3 seconds). Once all the numbers are plugged in, they are simplified down to find that the average force exerted upon the tennis ball is -2933.33 N (the negative sign accounts for the direction of the ball).
To sum up, the tennis ball was found to have a change in momentum of -17.6 kgm/s (indications of the subtraction sign would be that the tennis ball is traveling back towards the thrower). The contact between the ball and the wall was found to have an exertion of force upon the ball of -2933.33N.
Problem#2:
A 2.00 kg ball of clay was thrown at Kim and gets caught in her hair. Kim flies off of her chair at a speed of 0.160 m/s. How much does Kim weigh?
Equations and explanations:
Ball of Clay: Kim + clay:
m = 2.00 kg v = 4.0 m/s m = ? v = .160 m/s
PA + PB = PA+B
(2.00 kg*4.0 m/s + (x*0) = (.160 m/s) (200 kg + x)
8 kgm/s = (.160 m/s)(200 kg +x)
50 kg = 2.00 kg + x
x* = 48 kg
The Equation of Momentum Conservation
Plug in variables with x as a substitute for Kim's mass. Simplify the right side, and then divide both sides by (.160m/s). Simplify/solve.
Kim weighs 48 kg as shown above.
*x represents Kim's weight.
m = 2.00 kg v = 4.0 m/s m = ? v = .160 m/s
PA + PB = PA+B
(2.00 kg*4.0 m/s + (x*0) = (.160 m/s) (200 kg + x)
8 kgm/s = (.160 m/s)(200 kg +x)
50 kg = 2.00 kg + x
x* = 48 kg
The Equation of Momentum Conservation
Plug in variables with x as a substitute for Kim's mass. Simplify the right side, and then divide both sides by (.160m/s). Simplify/solve.
Kim weighs 48 kg as shown above.
*x represents Kim's weight.