A superball with mass equal to 50 grams is dropped from a height of h=1.5m . It

Question

A superball with mass equal to 50 grams is dropped from a height of h=1.5m . It collides with a table, then bounces up to a height of 1.0m . The duration of the collision (the time during which the superball is in contact with the table) is t=15ms . In this problem, take the positive y direction to be upward, and use g= 9.8 m/s^2 for the magnitude of the acceleration due to gravity. Neglect air resistance.

(B)Find the y component of the momentum of the ball immediately after the collision, that is, just as it is leaving the table. (p,after,y=?)

Explanation / Answer

To solve this problem, you just need to use conservation of momentum, conservation of energy and that impluse J = Ft = mv

A) KE = PE = 1/2mv2 = mgh or v = 2gh = 2*9.8*1.5m = 5.42m/s

B) Same as with A) but with a different height v = 2*9.8*1 = 4.42m/s

c) From the equation above F = mv/t = 0.05 (4.42-5.42)/0.015 = 3.33N

d) .05kg * (4.42m/s - 5.42m/s) = 0.05kgm/s

e)1/2 0.05 * 4.422 - 1/2*0.05 * 5.422 = -0.246J

Hope that helps

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.