Question

A 2.50-g bullet, traveling at a speed of 425 m/s, strikes the wooden block of a ballistic pendulum, such as that in the picture. The block has a mass of 215 g. (a) Find the speed of the bullet/block combination immediately after the collision. (b) How high does the combination rise above its initial position?

Answer 1

Answer:

Explanation:

mass of bullet, m = 2.5 g

initial velocity of bullet, u = 425 m/s

mass of block, M = 215 g

(a) Let the speed of bullet block system is v

Use the conservation of momentum

m x u + M x 0  = (M + m) x v

2.5 x 425 = ( 215 + 2.5) x v

v = 4.89 m/s

(b) Let it rise upto height h

Use conservation of energy

Kinetic energy after collision = Potential energy

1/2 (M + m) v² = ( M+ m) x gx h

0.5 x 4.89 x 4.89 = 9.8 x h

h = 1.22 m

Answer 2

Answer:

4.88505 m/s

1.21629 m

Explanation:

$m_1$ = Mass of bullet = 2.5 g

$m_2$ = Mass of block = 215 g

$v_1$ = Velocity of bullet = 425 m/s

$v_2$ = Velocity of block

Here, the linear momentum is conserved

$m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{2.5\times 10^{-3}\times 425+0.215\times 0}{2.5\times 10^{-3}+0.215}\\\Rightarrow v=4.88505\ m/s$

The speed of the combination of mass is 4.88505 m/s

The energy in the system is conserved

$\dfrac{1}{2}(m_1+m_2)v^2=(m_1+m_2)gh\\\Rightarrow h=\dfrac{v^2}{2g}\\\Rightarrow h=\dfrac{4.88505^2}{2\times 9.81}\\\Rightarrow h=1.21629\ m$

The maximum height the combined mass will reach is 1.21629 m