A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the grou
1 answer:
The final velocity of the ball is +9.25 m/s
Explanation:
We can solve the problem by using the impulse theorem, which states that the impulse exerted on the ball (product of the force exerted and the duration of the collision) is equal to the change in momentum of the ball:
where
F is the force exerted
is the duration of the collision
m is the mass of the ball
v is the final velocity
u is the initial velocity
In this problem, we have
m = 0.060 kg
u = -32 m/s
F = +55 N
Therefore we can solve for v to find the final velocity:
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Answer:
v ’= 21.44 m / s
Explanation:
This is a doppler effect exercise that changes the frequency of the sound due to the relative movement of the source and the observer, the expression that describes the phenomenon for body approaching s
f ’= f (v + v₀) / (v-)
where it goes is the speed of sound 343 m / s, v_{s} the speed of the source v or the speed of the observer
in this exercise both the source and the observer are moving, we will assume that both have the same speed,
v₀ = v_{s} = v ’
we substitute
f ’= f (v + v’) / (v - v ’)
f ’/ f (v-v’) = v + v ’
v (f ’/ f -1) = v’ (1 + f ’/ f)
v ’= (f’ / f-1) / (1 + f ’/ f) v
v ’= (f’-f) / (f + f’) v
let's calculate
v ’= (3400 -3000) / (3000 +3400) 343
v ’= 400/6400 343
v ’= 21.44 m / s