Please, see if is correct (2)

Jul 2008
179
1
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:

\(\displaystyle pa=ma.v\)

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

\(\displaystyle pa=pb\)

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

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

\(\displaystyle 0,01v=1,6315\)

\(\displaystyle v=163,15 m/s\)
 
May 2008
17
15
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.
 
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