A 7.0 g bullet is fired horizontally into a 1.5 kg block of wood suspended by a thread. The bullet emerges from the block with a speed of 195 m/s, and the block rises to a maximum height of 12 cm.

Find the initial speed of the bullet.

First calculate the momentum of the block with the bullet hole in it. Ignore the mass loss of wood, since they provide no information on it.

0.007*Vb = Total Momentum after bullet passes through
= 195*.007 + sqrt(2*g*H)*1.5

Solve for Vb, the initial bullet velocity
H = 0.12 m
sqrt(2*g*H) is the block velocity after the bullet passes through but before it swings upwards.

À 7gram billet fired intio à 1.5kg palliatif pendules thé billet émerge in thé block WIth à speed 200m/s and thé block rises to maximum Wright of 12cm fond thé initial speed of thé billet

To find the initial speed of the bullet, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision in an isolated system.

In this case, the bullet is fired horizontally into the block, so the initial momentum is solely due to the bullet. The final momentum is shared between the bullet (receding with a speed of 195 m/s) and the block (moving upward).

Let's denote the initial speed of the bullet as "v" and the final velocity of the block as "vf". Since the block rises to a maximum height and comes to rest, we can assume its final velocity is zero.

Momentum before collision = Momentum after collision
(mass of bullet) × (initial velocity of bullet) = [(mass of bullet) × (final velocity of bullet)] + [(mass of block) × (final velocity of block)]

Given:
Mass of bullet (m1) = 7.0 g = 0.007 kg
Mass of block (m2) = 1.5 kg
Final velocity of bullet (v1f) = 195 m/s
Final velocity of block (v2f) = 0 m/s

0.007 kg × v = (0.007 kg × 195 m/s) + (1.5 kg × 0 m/s)
0.007 kg × v = 1.365 kg⋅m/s
v = 1.365 kg⋅m/s ÷ 0.007 kg
v ≈ 195.0 m/s

Therefore, the initial speed of the bullet is approximately 195.0 m/s.