# Please, see if is correct (2)

#### Apprentice123

A ballistic pendulum may be used to measure the speed of a projétil. It comprises a block of wood of 2.5 kg suspended by a string of mass negligible. A bullet of mass equal to 10 grams is fired against the block, and after getting stuck in the block with that rises to a height of 65 cm. What is the speed of firing the bullet?

resolution:

$$\displaystyle pa=ma.v$$

$$\displaystyle pb=(ma+mb).d$$

$$\displaystyle pa=pb$$

$$\displaystyle ma.v=(ma+mb).d$$

$$\displaystyle 0,01v=(0,01+2,5).0,65$$

$$\displaystyle 0,01v=1,6315$$

$$\displaystyle v=163,15 m/s$$

#### flyingsquirrel

Hello,
A ballistic pendulum may be used to measure the speed of a projétil. It comprises a block of wood of 2.5 kg suspended by a string of mass negligible. A bullet of mass equal to 10 grams is fired against the block, and after getting stuck in the block with that rises to a height of 65 cm. What is the speed of firing the bullet?

resolution:

$$\displaystyle pa=ma.v$$

$$\displaystyle pb=(ma+mb).d$$
Linear momentum is defined by $$\displaystyle \vec{p}=m \vec{v}$$ where $$\displaystyle \vec{v}$$ is a velocity, not a distance.
$$\displaystyle pa=pb$$
Newton's second law states that $$\displaystyle \frac{\mathrm{d}\vec{p}}{\mathrm{d}t}=\vec{F}$$ where $$\displaystyle \vec{F}$$ is the sum of the forces exerted on the block and the bullet. To say that $$\displaystyle p_a=p_b$$ you need to have $$\displaystyle \vec{F}=\vec{0}$$ and that's not true since the string exerts a force on the block. As we don't know much about this force, I suggest you use conservation of energy instead of conservation of momentum.

At the beginning of the experiment, the energy $$\displaystyle E_1$$ of the block and of the bullet is the sum of :

• the kinetic energy of the block : that's $$\displaystyle 0\text{ J}$$ since the block is at rest ;
• the kinetic energy of the bullet : $$\displaystyle \frac{1}{2}m_1v_1^2$$ where $$\displaystyle v_1$$ is the velocity of the bullet when it's fired ;
• the gravitational potential energy of the bullet and of the block : this is $$\displaystyle 0\text{ J}$$ if we let $$\displaystyle z=0$$ be the height of the block and of the bullet at the beginning of the experiment.

Using all of this, $$\displaystyle E_1=\frac{1}{2}m_1v_1^2$$.

Now let $$\displaystyle E_2$$ be the energy of the block and of the bullet when the block reaches the height $$\displaystyle z_0=65\text{ cm}$$. At this time, the velocity of the block is $$\displaystyle 0\text{ m/s}$$ (this assumes that 65cm is the maximum height reached by the block) and its gravitational potential energy is $$\displaystyle (m_1+m_2)gz_0$$ so $$\displaystyle E_2=(m_1+m_2)gz_0$$. Equating $$\displaystyle E_1$$ and $$\displaystyle E_2$$ should give you the value of $$\displaystyle v_1$$ you're looking.

Apprentice123

#### Apprentice123

I thought:

V1=56,55 m/s

My method was wrong

#### Apprentice123

Thank you very much