A golf ball is placed on a basketball. The pair is then dropped from rest. With
ID: 1997065 • Letter: A
Question
A golf ball is placed on a basketball. The pair is then dropped from rest. With a bit of practice, you can get the golf ball to stay on top of the basketball until the basketball hits the ground. The mass of the golf ball is 0.0459kg, and the mass of the basketball is 0.619kg. (a) If the balls are released from a height where the bottom of the basketball is at 0.701m above the ground, what is the absolute value of the basketball's momentum just before it hits the ground? (b) What is the absolute value of the momentum of the golf ball at this instant? (c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collision in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions? (d) How high, measured from the ground, will the golf ball bounce up after its collision with the basketball?Explanation / Answer
Here,
m1 = 0.0459 Kg
m2 = 0.619 Kg
a)
speed of basketball's , u = sqrt(2 * g * h)
u = sqrt(2 * 9.8 * 0.701)
u = 3.711 m/s
absolute value of basketball = m2 * u
absolute value of basketball = 0.619 * 3.711 Kg.m/s
absolute value of basketball = 2.294 Kg.m/s
b)
speed of golf ball , u = sqrt(2 * g * h)
u = sqrt(2 * 9.8 * 0.701)
u = 3.711 m/s
absolute momentum of golfball = m2 * u
absolute momentum of golfball = 3.711 * 0.0459
absolute momentum of golfball = 0.1703 Kg.m/s
c)
after the collision , let the speed of golf ball is v2
and speed of basketball is v1
Using conservation of momentum
0.619 * v1 + 0.0459 * v2 = -0.1703 + 2.294 ---(1)
v2 - v1 = 2 * 3.711
solving for v1 and v2
v1 = 2.68 m/s
v2 = 10.10 m/s
absolute value of golf ball = 10.10 * 0.0459
absolute value of golf ball = 0.464 Kg.m/s
d)
let the heught is h
h = 10.10^2/(2 * 9.8)
h = 5.2 m
the golf ball will bounce to 5.2 m
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