A bullet of mass m = 5.89 g is fired into a block of mass M = 3.09 kg that is initially at rest on the edge of a smooth table. (The table top is 0.83 m above the floor.) The bullet ends up stuck in the block, and together they fly off the table and fall to the floor.

If the bullet has an initial speed of 870 m/s, what distance D will the block travel horizontally before landing on the floor?

m•v= (m+M) •V,

V = m•v/(m+M) = 5.89•10^-3•870/(5.89•10^-3+3.09) =
=1.294 m/s.
h=g•t^2/2 ,
t = sqrt(2•h/g) = sqrt(2•0.83/9.8) =
= 0.412 s,
D = V•t = 1.294•0.412 = 0.53 m.

To find the distance D that the block will travel horizontally before landing on the floor, we need to consider the conservation of linear momentum. The initial momentum of the bullet-block system is equal to the final momentum of the same system.

The initial momentum of the bullet is given by:
p_initial = m_bullet * v_bullet

Where:
m_bullet = mass of the bullet = 5.89 g = 0.00589 kg
v_bullet = initial speed of the bullet = 870 m/s

Substituting the values:
p_initial = 0.00589 kg * 870 m/s = 5.1153 kg·m/s

Since the block is initially at rest, its initial momentum is zero:
p_initial_block = 0 kg·m/s

The final momentum of the bullet-block system is given by:
p_final = (m_bullet + M_block) * v_final

Where:
v_final = final velocity of the bullet-block system

Since the bullet ends up stuck in the block, both the bullet and the block will have the same final velocity.

Now, to find the final velocity of the bullet-block system, we can use the principle of conservation of mechanical energy. The gravitational potential energy lost by the system while falling from a height h will be equal to the increase in kinetic energy of the system.

The gravitational potential energy lost by the system is given by:
PE_loss = m_total * g * h

Where:
m_total = total mass of the bullet-block system = m_bullet + M_block
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = height = 0.83 m (height of the table top above the floor)

The increase in kinetic energy of the system is given by:
KE_gain = (1/2) * (m_total) * v_final^2

According to the principle of conservation of mechanical energy, the gravitational potential energy lost is equal to the increase in kinetic energy:
PE_loss = KE_gain

Substituting the values and rearranging the equation:
m_total * g * h = (1/2) * (m_total) * v_final^2

Canceling out the common factor of m_total on both sides:
g * h = (1/2) * v_final^2

Solving for v_final:
v_final = sqrt(2 * g * h)

Substituting the values:
v_final = sqrt(2 * 9.8 m/s^2 * 0.83 m) = 3.4386 m/s

Since the final velocity of the system is the same as the final velocity of the block, we can now calculate the distance D that the block will travel horizontally before landing on the floor using the equation for horizontal motion:

D = v_final * t

Where:
D = horizontal distance traveled by the block
v_final = final velocity of the block (which we just found to be 3.4386 m/s)
t = time taken for the block to reach the floor

To find t, we can use the equation for vertical motion:

h = (1/2) * g * t^2

Where:
h = height = 0.83 m (height of the table top above the floor)
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time taken for the block to reach the floor

Rearranging the equation and solving for t:
t = sqrt((2 * h) / g)

Substituting the values:
t = sqrt((2 * 0.83 m) / 9.8 m/s^2) = 0.4128 s

Finally, substituting the values of v_final and t into the equation for D:
D = 3.4386 m/s * 0.4128 s = 1.4195 m

Therefore, the block will travel a horizontal distance of approximately 1.4195 meters before landing on the floor.