Part A A tennis ball bounces on the floor three times. If each time it loses 11%

Question

Part A

A tennis ball bounces on the floor three times. If each time it loses 11% of its energy due to heating, how high does it rise after the third bounce, provided we released it 3.3 m from the floor?

A block sliding along a horizontal frictionless surface with speed v collides with a spring and compresses it by 1.8 cm .

What will be the compression if the same block collides with the spring at a speed of 2v?

Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

after the first collision ,it has energy 0.89*m*g*h1 = 0.89*m*9.81*3.3 = 28.81*m

after first collision it goes to a height of m*g*h2 = 28.81*m

h2 = 28.81/9.81 = 2.94 m

after the second collision ,the ball energy 0.89*m*g*h2 = 0.89*m*9.81*2.94 = 25.67*m

after second collision it goes to a height h3 then

h3 = 25.66/9.81 = 2.61 m

after third collision,the ball has energy 0.89*m*9.81*2.61 = 22.84*m

after the third collision,the height reachd by the ball is h = 22.84/9.81 = 2.32 m = 232 cm = 230 cm

So the correct answer is 2) 230 cm

----------------------------------------------------------------

Apply law of conservation of energy

0.5*m*v^2 = 0.5*k*x^2

v1/v2 = x1/x2

v/2v = 1.8/x2

x2= 1.8*2 = 3.6 cm

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

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