A bullet with mass 4.95 g is fired horizontally into a 1.994 kg block attached to a horizontal spring. The spring has a constant 5.58 10^2 N/m and reaches a maximum compression of 6.27 cm.
(a) Find the initial speed of the bullet-block system.
(b) Find the speed of the bullet.
m = 0.005 kg
M = 2.000 kg
k = 6.00×10^2 = 600 N/m
x = 0.06 m
To find the initial speed of the bullet-block system, we can use the principle of conservation of momentum. In a collision, the total momentum before the collision is equal to the total momentum after the collision.
Let's assume the initial speed of the bullet is v and the final speed of the bullet-block system is vf.
(a) Find the initial speed of the bullet-block system:
1. Convert the mass of the bullet from grams to kilograms: 4.95 g = 0.00495 kg.
2. Apply the conservation of momentum equation: m1v1 = (m1 + m2)vf, where m1 is the mass of the bullet and m2 is the mass of the block.
(0.00495 kg)(v) = (0.00495 kg + 1.994 kg)(vf)
Simplify the equation: v = (0.00495 kg + 1.994 kg)(vf) / 0.00495 kg
3. Substitute the given values: v = (0.00495 kg + 1.994 kg)(vf) / 0.00495 kg
v = (1.99895 kg)(vf) / 0.00495 kg
v ≈ 403.8 vf
(b) Find the speed of the bullet:
Since the bullet is fired into the block and they are connected, the final velocity of the block and the bullet will be the same.
1. Apply the conservation of momentum equation for the bullet:
m1v1 = m1vf
(0.00495 kg)(v) = (0.00495 kg)(vf)
2. Substitute the known values: (0.00495 kg)(v) = (0.00495 kg)(vf)
3. Divide both sides of the equation by (0.00495 kg): v = vf
Therefore, the speed of the bullet is equal to the final velocity of the bullet-block system and can be found using part (a) as vf ≈ 403.8 vf.