1. Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via ele
1 answer:
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Answer:
x = 0.396 m
Explanation:
The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is spring
Data the putty has a mass m1 and velocity vo1, the block has a mass m2 . t's start using the moment to find the system speed.
Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash
p₀ = m1 v₀₁
Moment after shock
= (m1 + m2)
p₀ =
m1 v₀₁ = (m1 + m2)
= v₀₁ m1 / (m1 + m2)
= 4.4 600 / (600 + 500)
= 2.4 m / s
With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring
Before compressing the spring
Em₀ = K = ½ (m1 + m2) ²
After compressing the spring
= Ke = ½ k x²
As there is no rubbing the energy is conserved
Em₀ =
½ (m1 + m2) ² = = ½ k x²
x = √ (k / (m1 + m2))
x = 2.4 √ (11/3000)
x = 0.396 m