# Solving the Basic Mechanics Bullet Block Collision Problem

• Arm
In summary, the block and bullet move 1134 meters. The coefficient of kinetic friction between the block and surface of the table is 0.2.
Arm
Homework Statement
A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block which is
resting on a horizontal table. The bullet stops in the block and the block and bullet
combination move 2.0 m. The coefficient of kinetic friction between the block and
surface of the table is 0.2. Find the initial speed of the bullet.
Relevant Equations
$$KE = \frac{mv^2}{2}$$
$$W = \Delta E = F * distance$$
$$F_friction \le F_N * \mu$$
$$g = 10$$
$$\frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2)$$
v = 56.64
I just don't get how this is the wrong answer....it's just simple conservation of energy, right?

Arm said:
Homework Statement: A bullet with a mass of 5.0 g is fired horizontally into a 2.0 kg wooden block which is
resting on a horizontal table. The bullet stops in the block and the block and bullet
combination move 2.0 m. The coefficient of kinetic friction between the block and
surface of the table is 0.2. Find the initial speed of the bullet.
Relevant Equations: $$KE = \frac{mv^2}{2}$$
$$W = \Delta E = F * distance$$
$$F_friction \le F_N * \mu$$
$$g = 10$$

$$\frac{5E-3*v^2}{2} = (2 + 5E-3)(10)(0.2)(2)$$
v = 56.64
I just don't get how this is the wrong answer....it's just simple conservation of energy, right?
Energy is not conserved. The bullet deforms the block and generates heat in the process. It an inelastic collision. Also, friction is doing non-conservative work on the block. Linear Momentum is conserved immediately before and after the collision.

Arm
erobz said:
Energy is not conserved. The bullet deforms the block and generates heat in the process. It an inelastic collision. Also, friction is doing non-conservative work on the block. Linear Momentum is conserved immediately before and after the collision.
$$(5E-3)(v_1)=(5E-3+2)(v_2)$$
$$\frac{1}{2} (5E-3+2)({v_2}^2)=(5E-3+2)(10)(0.2)(2)$$
$$v_2 = 2.83$$
$$(5E-3)(v_1)=(5E-3+2)(2.83)$$
$$v_1 = 1134$$
thanks

erobz
Arm said:
$$(5E-3)(v_1)=(5E-3+2)(v_2)$$
$$\frac{1}{2} (5E-3+2)({v_2}^2)=(5E-3+2)(10)(0.2)(2)$$
$$v_2 = 2.83$$
$$(5E-3)(v_1)=(5E-3+2)(2.83)$$
$$v_1 = 1134$$
thanks
For future reference that notation is difficult to parse. Just leave it all variables next time. I kept thinking you were taking 5 times some quantity ##E## subtracting random numbers!

Also final numerical answers without units are meaningless. If I told you that I make two million a year, you might think I live on easy street. Not so fast. If the units of the two million is Ugandan shillings, that translates to \$500.00 USD waaaay below the poverty line.

erobz

## What is the basic setup of the bullet block collision problem?

The basic setup involves a bullet of mass $$m$$ fired horizontally into a block of mass $$M$$ that is initially at rest. The block is typically placed on a frictionless surface or suspended as a pendulum. The problem often requires determining the final velocities of the bullet and block after the collision, or the height to which the block rises if it swings upward.

## What principles are used to solve the bullet block collision problem?

The main principles used are the conservation of momentum and, in some cases, the conservation of energy. For an inelastic collision where the bullet embeds in the block, momentum conservation is key. If the problem involves the block rising to a certain height, energy conservation principles are also applied.

## How do you apply the conservation of momentum to this problem?

To apply the conservation of momentum, set the initial momentum of the system (bullet plus block) equal to the final momentum. The equation is $$m \cdot v_{\text{bullet initial}} = (m + M) \cdot v_{\text{final}}$$, where $$v_{\text{bullet initial}}$$ is the initial velocity of the bullet and $$v_{\text{final}}$$ is the final velocity of the combined bullet-block system.

## How do you determine the height to which the block rises after the collision?

After finding the final velocity of the bullet-block system using momentum conservation, use energy conservation to determine the height. The kinetic energy of the system just after the collision is converted into potential energy at the peak of the block's swing. Use the equation $$\frac{1}{2} (m + M) v_{\text{final}}^2 = (m + M) g h$$, where $$g$$ is the acceleration due to gravity and $$h$$ is the height.

## What assumptions are typically made in the bullet block collision problem?

Common assumptions include a frictionless surface or pivot, a perfectly inelastic collision (the bullet embeds in the block), and negligible air resistance. Additionally, it is often assumed that the mass of the bullet is much smaller than the mass of the block, simplifying some calculations.

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