Two forces 5N and 10N are acting at "O" and "P" respectively on a uniform rod of length 100 cm suspended at the position of cent
er of gravity 50cm mark as shown in figure .
1 answer:
Answer:
Option C. 75 cm
Explanation:
To obtain the position of P, let us calculate the value of y as shown in the attached photo.
The value of y can be obtained as follow:
Anticlockwise moment = clockwise moment
Anticlockwise moment = 5 × 50
Anticlockwise moment = 250
Clockwise moment = y × 10
Anticlockwise moment = clockwise moment
250 = y × 10
Divide both side by 10
y = 250/10
y = 25 cm
Finally, we shall obtain the value of P as follow:
Since P lies after the pivot (i.e 50 cm), therefore,
P = 50 + y
y = 25
P = 50 + 25
P = 75 cm
Therefore, the position of P on the metre rod is 75 cm
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Answer:
v₀ = 292.3 m / s
Explanation:
Let's analyze the situation, on the one hand we have the shock between the bullet and the block that we can work with at the moment and another part where the assembly (bullet + block) compresses a spring, which we can work with mechanical energy, as the data they give us are Let's start with this second part.
We write the mechanical energy when the shock has passed the bodies
Em₀ = K = ½ (m + M) v²
We write the mechanical energy when the spring is in maximum compression
=
= ½ k x²
Em₀ =
½ (m + M) v² = ½ k x²
Let's calculate the system speed
v = √ [k x² / (m + M)]
v = √[154 0.83² / (0.012 +0.104) ]
v = 30.24 m / s
This is the speed of the bullet + Block system
Now let's use the moment to solve the shock
Before the crash
p₀ = m v₀
After the crash
= (m + M) v
The system is formed by the bullet and block assembly, so the forces during the crash are internal and the moment is preserved
p₀ =
m v₀ = (m + M) v
v₀ = v (m + M) / m
let's calculate
v₀ = 30.24 (0.012 +0.104) /0.012
v₀ = 292.3 m / s