Question

A ballistic pendulum is used to measure the speed of high-speed projectiles. A bullet A weighing 22 g is fired into a 2.6-kg wood block B suspended by a cord of length l = 2.2 m. The block then swings through a maximum angle of θ = 60°. Determine the initial speed of the bullet v0.

Answer 1

Answer:

$v_{0}=554.19 m/s$

Explanation:

We can use the conservation of momentum and energy.

Conservation of momentum

$m_{1}v_{0}=(m_{1}+m_{2})V$ (1)

Where:

• m(1) is the mass of the bullet
• m(2) is the mass of the block
• v(0) is the initial speed of the bullet
• V(f) is the final speed of the system

Conservation of energy

We have kinetic energy at the beginning and we just have potential energy at the end because the block plus the bullet swing trough a maximum angle, it means they are at rest at that point.

$\frac{1}{2}(m_{1}+m_{2})V^{2}=(m_{1}+m_{2})gh$ (2)

The height h will be:

$h=L-Lcos(60)=L(1-cos(60))$

h is the difference height between the initial and final positions.

So, using the equation (2) we can find V.

$V^{2}=2gh$

$V=\sqrt{2gL(1-cos(60))}$

$V=\sqrt{2*9.81*2.2*(1-cos(60))}$

$V=4.65 m/s$

Finally, putting this value in the equation (1), the v(1i) will be:

$v_{0}=V\frac{(m_{1}+m_{2})}{m_{1}}$

$v_{0}=4.65\frac{(0.022+2.6)}{0.022}$

$v_{0}=554.19 m/s$

I hope it helps you!

Answer 2

Answer: initial velocity of bullet = 50.72m/s

Explanation:

Detailed explanation and calculation is shown in the image below.

NB: the wooden block has an initial velocity u = 0 since it was at rest.