Inelastic collisions fired bullet

[khawar]

A 5.5 gram bullet is fired into a block of wood with a mass of 22.6 grams. The wood block is initially at rest on a 1.5 meter tall post. After the collision the wood block and the bullet land 2.5 meters away from the base of the post. Find the initial speed of the bullet.
(After collision the bullet and the wood become one object) (The initial velocity is the velocity of the bullet when it strikes the wood)

 

[Doc Al]

This probably belongs in the homework help section. (And no calculus needed.)

Show what you've done so far.

Hint: During the collision, momentum is conserved. After the collision, the combined body is a projectile.

 

[M98Ranger]

To find the initial speed of the bullet, we can use the conservation of momentum principle. In an inelastic collision, the total momentum before and after the collision remains the same.

Let us define our variables:
m1 = mass of bullet = 5.5 grams = 0.0055 kg
m2 = mass of wood block = 22.6 grams = 0.0226 kg
v1 = initial velocity of bullet
v2 = final velocity of combined object

Applying the conservation of momentum principle, we can write the equation as:
m1v1 = (m1 + m2)v2

Substituting the values, we get:
0.0055 kg * v1 = (0.0055 kg + 0.0226 kg) * v2
0.0055 kg * v1 = 0.0281 kg * v2

Now, we need to find the final velocity of the combined object. We can use the equation of motion, s = ut + 1/2at^2, where s is the displacement, u is the initial velocity, a is the acceleration and t is the time taken.

Here, s = 2.5 m, u = 0 m/s (since the wood block and bullet were initially at rest), a = 9.8 m/s^2 (due to gravity), t = time taken for the combined object to travel 2.5 m

Substituting the values, we get:
2.5 m = 0 + 1/2 (9.8 m/s^2) * t^2
2.5 m = 4.9 m/s^2 * t^2
t^2 = 2.5 m / 4.9 m/s^2 = 0.5102 s
t = √0.5102 s = 0.714 s

Now, we can find the final velocity, v2, using the equation of motion:
s = ut + 1/2at^2
2.5 m = 0 + 1/2 (9.8 m/s^2) * (0.714 s)^2
2.5 m = 0.5 * 4.9 m/s^2 * 0.5102 s^2
v2 = 2.5 m /