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.

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

answer: the correct answer id option(d).

explanation:

snell's law refraction:

the angle of refraction =

the angle of incidence =

refractive index of air = = 1

refractive index of glycerol = 1.47

from the given options, the most closest answer to 63.72° is 65° of option (d).