24. A bullet of mass m = 15gm is fired with an initial speed of 50om/s into a block of mass
M = 0.8kg initially at rest at the edge of a frictionless table of height h. The bullet
remains in the block, and after impact the block lands a distance d from the bottom of the
table. Determine the initial speed of the bullet and the distance d at which the block hits
the floor.
To solve this problem, we need to apply the principles of conservation of momentum and conservation of energy.
1. Conservation of momentum:
Before the collision, the bullet and the block are separate, so the total initial momentum is zero. After the collision, they are stuck together, so we have:
(Initial momentum of bullet) + (Initial momentum of block) = 0
(m x v) + 0 = (m + M) x V
where:
m = mass of the bullet (15 gm = 0.015 kg)
v = initial speed of the bullet (unknown)
M = mass of the block (0.8 kg)
V = velocity of the combined bullet and block after the collision (unknown)
2. Conservation of energy:
The total initial energy (kinetic energy) of the bullet will be equal to the sum of the energy transferred to the block, gravitational potential energy, and kinetic energy of the combined bullet and block at the bottom. We can write the equation as follows:
(Initial kinetic energy of bullet) = (Energy transferred to block) + (Gravitational potential energy at height h) + (Final kinetic energy of bullet and block)
(1/2) m v^2 = (1/2) (m + M) V^2 + M g h + (1/2) (m + M) V^2
where:
g = acceleration due to gravity (9.8 m/s^2)
h = height of the table (unknown)
Now let's solve the equations to find the unknowns.
From the first equation:
m x v = (m + M) x V
Substituting the known values:
0.015 kg x v = (0.015 kg + 0.8 kg) x V
v = (0.815 kg / 0.015 kg) x V
v = 54.33V
Now let's substitute this value of v into the second equation.
From the second equation:
(1/2) m v^2 = (1/2) (m + M) V^2 + M g h + (1/2) (m + M) V^2
Substituting the known values:
(1/2) x 0.015 kg x (54.33V)^2 = (1/2) (0.015 kg + 0.8 kg) V^2 + 0.8 kg x 9.8 m/s^2 x h + (1/2) (0.015 kg + 0.8 kg) V^2
Simplifying the equation:
0.004667 x V^2 = 0.0008075 x V^2 + 7.84 x h + 0.0008075 x V^2
Rearrange the equation:
7.84 x h = 0.004667 x V^2
Since the bullet and block remain stuck together, they will hit the ground at the same time. At the point of impact, the vertical distance the block travels is equal to the height of the table. Therefore, the distance d at which the block hits the floor is equal to the height of the table, h.
To calculate the value of height (h), we need to rearrange the equation:
h = (0.004667 x V^2) / 7.84
Now we can substitute the initial speed of the bullet (v) into the equation to find the height:
h = (0.004667 x (54.33V)^2) / 7.84
Simplifying the equation will give you the value of h.
Finally, the initial speed of the bullet can be found by substituting the value of v calculated previously:
v = (0.815 kg / 0.015 kg) x V
Solving for v will give you the initial speed of the bullet.
Please note that the final calculations require substituting the values of V and h into the equations to obtain the final answers.