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g A 10.0 g bullet is fired into and embeds itself in a 1.65 kg block attached to a spring with a spring constant of 17.2 N/m and whose mass
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g A 10.0 g bullet is fired into and embeds itself in a 1.65 kg block attached to a spring with a spring constant of 17.2 N/m and whose mass is negligible. How far is the spring compressed if the bullet has a speed of 300 m/s just before it strikes the block, and the block slides on a frictionless surface? (Note: You must use conservation of momentum in this problem.
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3 years
2021-08-24T02:24:34+00:00
2021-08-24T02:24:34+00:00 1 Answers
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Answer:
the compression of the spring is 0.5613 m
Explanation:
Given;
mass of bullet, m₁ = 0.01 kg
mass of block, m₂ = 1.65 kg
initial velocity of the block, u₂ = 0
initial velocity of the bullet before hitting the block, u₁ = 300 m/s
Final speed of the bullet-block system after collision, v = ?
spring constant, K = 17.2 N/m
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v( m₁ + m₂)
0.01 x 300 + 1.65 x 0 = v(0.01 + 1.65)
3 = 1.66v
v = 3 / 1.66
v = 1.807 m/s
Apply the principle of conservation of mechanical energy to determine the compression of the spring;
KE₁ + PE₁ = KE₂ + PE₂
¹/₂mu² + ¹/₂Kx₁² = ¹/₂mv² + ¹/₂Kx₂²
where;
m is mass of bullet and block embedded together
u is the initial velocity of the bullet-block system = 1.807 m/s
v is the final velocity of the bullet-block system = 0
x₁ is the initial compression of the spring = 0
x₂ is the final compression of the spring = ?
¹/₂(1.65 + 0.01) (1.807)² + ¹/₂(17.2)(0)² = ¹/₂(1.65 + ).01)(0)² + ¹/₂(17.2)(x₂)²
2.71 + 0 = 0 + 8.6(x₂)²
(x₂)² = 2.71 / 8.6
(x₂)² = 0.3151
x₂ = √0.3151
x₂ = 0.5613 m
Therefore, the compression of the spring is 0.5613 m