A block of mass m 1 sits at rest on a table over a hole in the table. From below the table, a
bullet of mass m 2 is fired vertically upwards into the block. The bullet imbeds itself in the block.
The block and bullet rise to a maximum height h. What is the bullet’s initial speed? Answer in
terms of m 1, m 2, h, and g.
The law of conservation of linear momementum
m1•v +0=(m1+m2) •u
v=(m1+m2) •u/m1.
The law of conservation of energy
(m1+m2) •u²/2=(m1+m2) •g•h,
u=sqrt(2•g•h).
v=(m1+m2) •u/m1= (m1+m2) •sqrt(2•g•h)/ m1.
To find the bullet's initial speed, we can use the principle of conservation of mechanical energy. At the maximum height h, the system (block + bullet) will have maximum potential energy and zero kinetic energy.
Let's denote the bullet's initial speed as v₀.
First, let's find the potential energy (PE) at maximum height h. The potential energy is given by:
PE = m₁ * g * h
where m₁ is the mass of the block and g is the acceleration due to gravity.
Next, let's find the kinetic energy (KE) at maximum height h. The kinetic energy is given by:
KE = (1/2) * (m₁ + m₂) * v²
where m₂ is the mass of the bullet and v is its velocity.
Since the bullet imbeds itself in the block, the final velocity of the system at maximum height h is equal to zero. Therefore, we have:
KE = 0
Setting PE equal to KE, we get:
m₁ * g * h = (1/2) * (m₁ + m₂) * v₀²
Now, rearrange the equation to solve for v₀:
v₀² = (2 * m₁ * g * h) / (m₁ + m₂)
Taking the square root of both sides, we find:
v₀ = √((2 * m₁ * g * h) / (m₁ + m₂))
So, the bullet's initial speed in terms of m₁, m₂, h, and g is √((2 * m₁ * g * h) / (m₁ + m₂)).