Question

Two ball 'A' and 'B' each of mass 1.6kg and are rolling on a horizontal table
2m above the floor. Both roll off the end of the table and reach the floor
below. Ball 'A' has horizontal velocity of 4.0m/s as it leaves table, while 'B'
has a velocity of 2.5 m/s. What is the difference in horizontal distance
travelled by A and B before the strike the floor?
A0.6 m
B. 0.7 m
C. 0.8 m
D. 0.9 m
E. 1.0 m
wat is kinetic energy of ball B just before it
ERK​

Answer 1

Answer:

0.96m

Explanation:

Both balls had only horizontal velocity while leaving the table. So initially, vertical velocity of both the balls were 0.

height of table , h = 2m

time taken to reach the floor = √(2h/g) = √(2×2/9.8) = 0.64 s

horizontal distance travelled by the balls in 0.64s = horizontal speed × time

by A = 4 × 0.64 = 2.56m

by B = 2.5 × 0.64 = 1.60 m

difference = 2.56 - 1.60 = 0.96m

____________________________

for ball B,

mass = 1.6 kg

just before striking floor,

vertical velocity, v = v + at = 0 + 9.8×0.64 = 6.3 m/s

horizontal velocity, u = u0 + at = 2.5 + 0×0.64 = 2.5 m/s

velocity, V = √(v^2 + u^2) =√(6.3^2 + 2.5^2) = 6.8 m/s

kinetic energy = 1/2 m V^2 = 1/2 × 1.6 × 6.8^2 = 36.75 Joule

Answer 2

Answer:

Hey mate

Here your answer

0.96m

0.96m is the correct answer