Find the initial speed of the bullet

[vangjo]

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 from the base of the post. Find the initial speed of the bullet.

 

[Integral]

This is a 2 part problem. First you need to find the initial velocity of the block using the equations of motion for a falling body.

Find the time the block was falling with

$$\frac {-gt^2} 2 + V_y(0)+y(0)$$

Then assume that the block had its full x velocity before it started to drop so

$$X(t)= V_x t$$

Also use conservation of momentum

$$M_{bullet}V_{bullet}= M_{Block + Bullet} V_{Block}$$

That should get you started.

 

[M98Ranger]

To find the initial speed of the bullet, we can use the conservation of momentum principle, which states that the total momentum before a collision is equal to the total momentum after the collision. In this scenario, the bullet and the wood block are the only objects involved in the collision, so we can set up the following equation:

(mass of bullet)(initial velocity of bullet) = (mass of bullet + mass of wood block)(final velocity of bullet + final velocity of wood block)

Plugging in the given values, we get:

(5.5 g)(initial velocity) = (5.5 g + 22.6 g)(final velocity)

Simplifying, we get:

5.5 g(initial velocity) = 28.1 g(final velocity)

Dividing both sides by 5.5 g, we get:

initial velocity = (28.1 g(final velocity))/5.5 g

Now, to solve for the final velocity, we can use the formula for projectile motion:

final velocity = √(2gh)

Where g is the acceleration due to gravity (9.8 m/s²) and h is the height of the post (1.5 m).

Plugging in the values, we get:

final velocity = √(2(9.8 m/s²)(1.5 m)) = 5.42 m/s

Substituting this value into our initial velocity equation, we get:

initial velocity = (28.1 g(5.42 m/s))/5.5 g = 27.57 m/s

Therefore, the initial speed of the bullet is approximately 27.57 m/s.