A block of wood with a mass of 1300g is sitting on a horizontal tabletop. The coefficient of kinetic friction between the block and table is µk=0.74.A rifle bullet with a mass of 39g is fired horizontally into the block at a sped of 290m/s and stops inside the block.
a-what is the velocity of the bullet/block system after impact?
b-what is the acceleration/deceleration experienced by the bullet/block system due to the friction?
c-how far does the block with the bullet embedded in it slide before coming to a stop?
a.)
m1v1+m2v2=m3v3
Let:
m1=1300g=1.3kg
v1=0m/s
m2=39g=0.039kg
v2=290m/s
m3=1.339kg *** addition of m1 and m2
and
v3=???
Solve for v3:
(1.3kg+0)+(0.039kg+290m/s)=(1.339kg)v3
v3=[(1.3kg+0)+(0.039kg+290m/s)]/1.339kg
v3=1.3+11.31/1.339
v3=9.41m/s
b.)
Fnet=F(block)-F(friction)
Let
Fnet=m3*a
F(block)=m3*g=(1.339kg*9.8m/s^2)
**g=gravity=9.8m/s^2
and
F(friction)=m3*g*(0.74)
Fnet=13.12N-9.7N
Fnet=3.42N
3.42N=1.339kg*a
a=3.42N/1.339kg
a=2.55m/s^2
c.)
In order for the block to stop, the work done by kinetic friction must equal 0. That is, the force of friction does takes away all of the kinetic energy of the system.
0=Kinetic Energy-Work done by Friction
0=1/2mv^2-F*d
F*d=1/2mv^2
d=(1/2mv^2)/F
let
m=m3=1.339kg **From a
v=v3=9.41m/s ***From a
and
F=9.7N *** From b
Solve for d:
d=(0.5(1.339kg*9.41m/s^2)^2)/9.7N
d=6.12m
*** Hopefully, someone checks this.
To solve this problem, we can apply the principles of conservation of momentum, Newton's second law, and equations related to friction.
Step 1: Calculate the momentum of the bullet before impact.
The momentum of an object is given by the equation p = mv, where p is momentum, m is mass, and v is velocity.
Given:
Mass of the bullet (m1) = 39g = 0.039 kg
Velocity of the bullet before impact (v1) = 290 m/s
Using the equation, p1 = m1v1, we can calculate the momentum of the bullet before impact.
p1 = (0.039 kg)(290 m/s) = 11.31 kg·m/s
Step 2: Calculate the momentum of the block before impact.
Given:
Mass of the block (m2) = 1300g = 1.3 kg
Velocity of the block before impact (v2) = 0 m/s (since it is at rest)
Using the equation, p2 = m2v2, we can calculate the momentum of the block before impact.
p2 = (1.3 kg)(0 m/s) = 0 kg·m/s
Step 3: Calculate the total momentum of the system before impact.
The total momentum of an isolated system is conserved before and after an impact.
Total momentum before impact = momentum of bullet before impact + momentum of block before impact
p_initial = p1 + p2
p_initial = 11.31 kg·m/s + 0 kg·m/s
p_initial = 11.31 kg·m/s
Step 4: Calculate the velocity of the bullet/block system after impact.
Given:
Total mass of the bullet/block system (m_total) = m1 + m2 = 0.039 kg + 1.3 kg = 1.339 kg
Using the equation, p_final = m_total * v_final, we can find the velocity of the bullet/block system after impact.
v_final = p_final / m_total
Since the total momentum is conserved, p_initial = p_final
Therefore, v_final = p_initial / m_total
v_final = 11.31 kg·m/s / 1.339 kg
v_final ≈ 8.45 m/s
a) The velocity of the bullet/block system after impact is approximately 8.45 m/s.
Step 5: Calculate the frictional force acting on the bullet/block system.
The frictional force can be calculated using the equation, F_friction = µk * (normal force), where µk is the coefficient of kinetic friction and (normal force) is the force exerted perpendicular to the surface.
The normal force is equal to the weight (mg) of the block, where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
µk = 0.74
m2 = 1.3 kg
F_friction = µk * (normal force) = µk * (m2 * g)
F_friction = 0.74 * (1.3 kg * 9.8 m/s^2)
F_friction ≈ 11.46 N
b) The acceleration experienced by the bullet/block system due to friction can be calculated using Newton's second law, F = ma, where F is the net force and a is the acceleration.
Since the block is moving horizontally, the net force is the force of friction acting in the opposite direction.
F = - F_friction = - 11.46 N
m_total = 1.339 kg (mass of the bullet/block system)
Using the equation, F = ma, we can find the acceleration.
- 11.46 N = (1.339 kg) * a
a = - 11.46 N / 1.339 kg ≈ - 8.56 m/s^2
c) The distance the block with the bullet embedded in it slides before coming to a stop can be calculated using the equation, d = (v_final)^2 / (2 * |a|), where d is the distance, v_final is the final velocity, and |a| is the magnitude of the acceleration.
Using the given values of v_final ≈ 8.45 m/s and |a| ≈ 8.56 m/s^2, we can find the distance it slides.
d = (8.45 m/s)^2 / (2 * 8.56 m/s^2)
d ≈ 4.17 m
c) The block with the bullet embedded in it slides approximately 4.17 meters before coming to a stop.
To answer the given questions, we'll use the laws of motion and principles of friction. Let's break down each question step by step:
a) What is the velocity of the bullet/block system after impact?
The bullet and the block will form a system after impact, where their momenta will be equal. To calculate the velocity of the system, we'll use the principle of conservation of momentum.
The initial momentum of the bullet is given by its mass (m1) multiplied by its initial velocity (v1):
Momentum_before = m1 * v1
The initial momentum of the block is zero since it was initially at rest.
After impact, the bullet and the block move together. Let the final velocity of the bullet/block system be v2.
The final momentum of the system is the sum of the momentum of the bullet and the momentum of the block, which can be written as:
Momentum_after = (m1 + m2) * v2
According to the conservation of momentum, Momentum_before = Momentum_after.
m1 * v1 = (m1 + m2) * v2
Now we can solve for v2:
v2 = (m1 * v1) / (m1 + m2)
Substituting the given values:
m1 = 39g = 0.039kg
v1 = 290m/s
m2 = 1300g = 1.3kg
v2 = (0.039 * 290) / (0.039 + 1.3) = 0.11 m/s (approximately)
Therefore, the velocity of the bullet/block system after impact is 0.11 m/s.
b) What is the acceleration/deceleration experienced by the bullet/block system due to friction?
To find the acceleration or deceleration experienced by the bullet/block system due to friction, we can use Newton's second law of motion.
The force of friction (F_friction) can be calculated by multiplying the coefficient of kinetic friction (μk) with the normal force (F_normal) acting on the block. In this case, the normal force is equal to the weight of the block.
F_friction = μk * F_normal
The normal force (F_normal) can be calculated by multiplying the mass of the block (m2) with the acceleration due to gravity (g).
F_normal = m2 * g
The acceleration or deceleration experienced by the bullet/block system (acceleration) can be calculated using Newton's second law:
F_net = m_total * acceleration
Since the force of friction (F_friction) acts in the opposite direction to the motion, it should have a negative sign.
F_net = m_total * acceleration = -F_friction
Therefore, the acceleration can be written as:
acceleration = -F_friction / m_total
Substituting the given values:
μk = 0.74
m2 = 1300g = 1.3kg
g = 9.8m/s^2 (acceleration due to gravity)
F_normal = m2 * g = 1.3 * 9.8 = 12.74N (approximately)
F_friction = μk * F_normal = 0.74 * 12.74 = 9.43N (approximately)
m_total = m1 + m2 = 0.039 + 1.3 = 1.339kg (approximately)
acceleration = -F_friction / m_total = -9.43 / 1.339 ≈ -7.04 m/s^2
Therefore, the acceleration or deceleration experienced by the bullet/block system due to friction is approximately -7.04 m/s^2.
c) How far does the block with the bullet embedded in it slide before coming to a stop?
To find the distance traveled by the block with the bullet embedded in it, we'll use the equation of motion for uniformly accelerated motion.
The equation that relates distance (d), initial velocity (v1), acceleration (a), and final velocity (v2) can be written as:
v2^2 = v1^2 + 2ad
Rearranging the equation, we get:
d = (v2^2 - v1^2) / (2a)
Substituting the given values:
v1 = 290m/s
v2 = 0.11 m/s
a = -7.04 m/s^2 (since the system is decelerating due to friction)
d = (0.11^2 - 290^2) / (2 * -7.04)
Calculating the value of d will give you the distance traveled by the block with the bullet embedded in it.
d ≈ 12.45 m (approximately)
Therefore, the block with the bullet embedded in it slides approximately 12.45 meters before coming to a stop.