Explanation:
Adding all the weighted values on the right hand side, we get
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The distance traveled by the wood after the bullet emerges is 0.16 m.
The given parameters;
- <em>mass of the bullet, m = 23 g = 0.023 g</em>
- <em>speed of the bullet, u = 230 m/s</em>
- <em>mass of the wood, m = 2 kg</em>
- <em>final speed of the bullet, v = 170 m/s</em>
- <em>coefficient of friction, μ = 0.15</em>
The final velocity of the wood after the bullet hits is calculated as follows;
The acceleration of the wood is calculated as follows;
The distance traveled by the wood after the bullet emerges is calculated as follows;
Thus, the distance traveled by the wood after the bullet emerges is 0.16 m.
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Answer:
0.661 s, 5.29 m
Explanation:
In the y direction:
Δy = 2.14 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
(2.14 m) = (0 m/s) t + ½ (9.8 m/s²) t²
t = 0.661 s
In the x direction:
v₀ = 8 m/s
a = 0 m/s²
t = 0.661 s
Find: Δx
Δx = v₀ t + ½ at²
Δx = (8 m/s) (0.661 s) + ½ (0 m/s²) (0.661 s)²
Δx = 5.29 m
Round as needed.