In: Physics
I have the answers listed I just want to know if someone could show me the work step by step!
Q1. A ball of mass 60 g is dropped from a height of 3.4 m. It lands on the top of a frictionless ramp at height 1.8 m. The ramp is tilted at an angle of 20 degrees.
(a) What is the velocity of the ball at the top of the ramp? Answer: 5.6 m/s
(b) At the bottom of the ramp it collides with and sticks to a ball of mass 73 g. What is their velocity after the collision? Answer: 3.68 m/s
(c) The stuck together balls collide with a spring of spring constant 300 N/m. How much will they compress it? Answer: 0.077m
(d) They then go back up the ramp. How high will they go? Answer: 0.7 m
a) height dropped by the first ball = h = (3.4 - 1.8) = 1.6 m
when this ball hits the ramp , it's speed is u = [answer]
b) height of ramp = 1.8 m
length of inclined plane =
speed of the first ball when it reaches the bottom of the ramp = v =
this is the velocity of first ball before collision.
they stick to each other after collision.
v' = velocity of the bodies after collision.
Now applying conservation of linear momentum :
60 x 8.16 = (60 + 73) x v'
=> v' = 3.68 m/s [answer]
c) this system now moves forward and compresses a spring of spring constant k = 300 N/m say by x.
from conservation of mechanical energy:
The loss in kinetic energy of the balls = gain in potential energy of the spring
=> [answer]
d) then again after compression, the spring relaxes and the balls moves again upward.
From conservation of mechanical energy, the balls gains the same speed with which it hits the spring.
that is , v' = 3.68 m/s.
with this speed the balls moves upward but by retardation due to gravity.
therefore, using kinematic equation :
= 0.7 m [answer]