A bullet of mass m is fired at speed v0 into a wooden block of mass M. The bullet instantaneously comes to rest in the block. The block with the embedded bullet slides along a horizontal surface with a coefficient of kinetic friction . What is the algebraic expression which determines how far the block slides before it comes to rest (the magnitude of displacement s in the figure)?
To determine the algebraic expression that determines how far the block slides before it comes to rest, we need to apply the principle of conservation of momentum and consider the forces acting on the system.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant if no external forces are acting on it. In this case, the bullet and the block form a closed system, neglecting external forces like air resistance.
Initially, the bullet has momentum p1 = m * v0. Since the bullet comes to rest in the block, the final momentum of the bullet and the block system is zero (p2 = 0).
According to the principle of conservation of momentum:
p1 + 0 = p2
m * v0 = (m + M) * v
Where v is the velocity of the bullet and block system after the collision.
To determine the velocity v of the block and bullet system after the collision, we need to consider the forces acting on it.
The only external force acting horizontally on the block is the force of kinetic friction (fk) given by:
fk = μ * N,
where μ is the coefficient of kinetic friction and N is the normal force acting on the block. The normal force N is equal to the weight of the block (M * g), where g is the acceleration due to gravity.
Now, the force of kinetic friction can be written as:
fk = μ * M * g
The force of kinetic friction acts on the block in the direction opposite to its motion, so we can write the equation of motion for the block as:
fk = (M + m) * a,
where a is the acceleration of the block.
Since the block slides until it comes to rest (final velocity is zero), we can use the equation of motion for constant acceleration to find the displacement (s) of the block:
v^2 = u^2 + 2a * s,
where u is the initial velocity of the block, which is the same as the velocity v of the bullet and block system after the collision.
Since we know the initial velocity u is equal to v0, and we also have the equation fk = (M + m) * a, we can substitute these values into the equation of motion to solve for the displacement s.
Hence, the algebraic expression that determines how far the block slides before coming to rest is:
v^2 = v0^2 + 2 * μ * M * g * s,
where v is the velocity of the bullet and block system after the collision, μ is the coefficient of kinetic friction, M is the mass of the wooden block, g is the acceleration due to gravity, and s is the displacement of the block.
To determine the algebraic expression for the displacement of the block before it comes to rest, we need to consider the work done on the block by the friction force.
The work done on an object by a force is defined as the product of the force applied and the displacement of the object in the direction of the force. In this case, the friction force is the only force acting on the block, and it causes it to come to rest.
The work done by the friction force is given by the formula:
Work = Force * Displacement * cos(theta)
Since the friction force is acting horizontally and in the opposite direction of motion, the angle between the force and displacement is 180 degrees, making cos(theta) = -1.
The force of friction can be calculated using the formula:
Force of friction = coefficient of kinetic friction * Normal force
The normal force (N) is equal to the weight of the block (Mg), where g is the acceleration due to gravity.
Therefore, the force of friction is:
Force of friction = μ * M * g
Now, we can substitute the values of the force of friction and the angle into the work formula:
Work = (μ * M * g) * Displacement * (-1)
Since the bullet instantaneously comes to rest in the block, we can assume the work done on the block by the friction force is equal to the kinetic energy of the bullet before it is embedded.
The initial kinetic energy of the bullet is:
Kinetic energy = (1/2) * m * v0^2
Equating the work done by friction to the kinetic energy, we have:
(μ * M * g) * Displacement * (-1) = (1/2) * m * v0^2
Simplifying the equation for displacement (s):
Displacement (s) = -(m * v0^2) / (2 * μ * M * g)
Therefore, the algebraic expression for the displacement of the block before it comes to rest is:
s = -(m * v0^2) / (2 * μ * M * g)
momentumbefore=momentumafter
mVo=(m+M)V
solve for V of the combo.
Initial KE=friction work
1/2 (m+M)V^2=mu(g)(m+M)*d
solve for d.
d= 1/2 (m/(m+M))^2Vo^2/mu*g
check that.