A steel ball of mass m=5 g is moving at a speed of 250 m/s toward a large mass M=2.2 kg which is connected to a relaxed spring, as shown below. When the ball reaches the large mass it collides and bounces off without losing any kinetic energy. The large mass is connected to a spring of spring constant k=53 N/m and after the collision the large mass moves to the left, compressing the spring. The mass slides on a frictional surface with kinetic coefficient of friction 0.80. Find the distance d by which the spring is compressed when the large mass comes briefly to rest. d=?
First, let's find the initial momentum of the steel ball and the large mass. The momentum of the steel ball is $p_{b, initial} = m_bv_b = (5 \times 10^{-3}\,\text{kg})(250\,\text{m/s}) = 1.25\,\text{kg m/s}$. Since the large mass is initially at rest, its initial momentum is zero.
After the collision, the steel ball and the large mass will have the same final momentum since no kinetic energy is lost. Let $v_{b, final}$ and $v_{M, final}$ be the final velocities of the steel ball and the large mass, respectively. The conservation of momentum equation is:
$$p_{b, initial} + p_{M, initial} = p_{b, final} + p_{M, final}$$
Since the initial momentum of the large mass is zero, the equation simplifies to:
$$1.25\,\text{kg m/s} = m_bv_{b, final} + Mv_{M, final}$$
Now, let's use the conservation of energy to find the relationship between the final velocities of the steel ball and the large mass. Half of the sum of the initial and final kinetic energies of the steel ball is stored as potential energy in the spring when it is compressed:
$$\frac{1}{2}m_b(v_{b, initial}^2 + v_{b, final}^2) = \frac{1}{2}k d^2$$
Since we are given that the steel ball's kinetic energy does not change, $v_{b, initial} = v_{b, final}$. Substituting this into the equation:
$$\frac{1}{2}(5 \times 10^{-3}\,\text{kg})(250\,\text{m/s})^2 = \frac{1}{2}(53\,\text{N/m}) d^2$$
Solving for $d^2$:
$$d^2 = \frac{(5 \times 10^{-3}\,\text{kg})(250\,\text{m/s})^2}{53\,\text{N/m}}$$
$$d^2 = 0.147\,\text{m}^2$$
Therefore, the distance $d = \sqrt{0.147\,\text{m}^2} = 0.383\,\text{m}$ (to three significant figures). The spring is compressed by 0.383 meters when the large mass comes briefly to rest.
To find the distance by which the spring is compressed when the large mass comes to rest, we need to follow these steps:
Step 1: Calculate the initial momentum of the steel ball before the collision.
Step 2: Calculate the initial velocity of the large mass after the collision using the principle of conservation of momentum.
Step 3: Calculate the deceleration of the large mass due to friction.
Step 4: Calculate the distance the large mass travels until it comes to rest using the equation of motion.
Step 5: Calculate the displacement of the spring when the large mass comes to rest using Hooke's Law.
Let's proceed with each step one by one:
Step 1: Calculate the initial momentum of the steel ball before the collision.
Momentum (p) = mass (m) × velocity (v)
p = 5 g × 250 m/s
Note: We need to convert the mass to kg and the momentum to kg·m/s.
Since 1 g = 0.001 kg and 1 kg·m/s = 1 N·s, we have:
p = 0.005 kg × 250 m/s
p = 1.25 kg·m/s
Step 2: Calculate the initial velocity of the large mass after the collision using the principle of conservation of momentum.
According to the principle of conservation of momentum:
Momentum before collision = Momentum after collision
Since the steel ball bounces back without losing kinetic energy, the initial momentum of the large mass is equal in magnitude to the momentum of the steel ball.
Therefore, the initial velocity of the large mass (v') is given by:
v' = (Initial momentum of the steel ball) / (Mass of the large mass)
v' = (-1.25 kg·m/s) / (2.2 kg)
Note: The negative sign indicates that the velocity is in the opposite direction.
Step 3: Calculate the deceleration of the large mass due to friction.
The frictional force (f) is given by:
f = coefficient of kinetic friction × normal force
The normal force (N) is equal to the weight of the large mass:
N = Mass of the large mass × gravity
N = 2.2 kg × 9.8 m/s^2
Now, the frictional force (f) can be calculated as:
f = 0.80 × (2.2 kg × 9.8 m/s^2)
The acceleration of the large mass (a) due to friction can be calculated using Newton's second law:
f = mass × acceleration
a = f / mass
a = (0.80 × (2.2 kg × 9.8 m/s^2)) / 2.2 kg
Step 4: Calculate the distance the large mass travels until it comes to rest using the equation of motion.
The equation of motion can be used to find the distance traveled (d) by the large mass until it comes to rest:
v'^2 = v^2 + 2as
0 = (v')^2 + 2(-a)s
Step 5: Calculate the displacement of the spring when the large mass comes to rest using Hooke's Law.
Hooke's Law states that the force exerted by a spring is proportional to its displacement:
F = -kx
In this case, the force exerted by the spring (F) is equal to the frictional force (f) due to the mass coming to rest. Therefore:
f = -kx
Since the spring pushes the mass in the opposite direction, we can use the magnitude of f, so:
|-f| = kx
|x| = |-f| / k
Finally, the distance by which the spring is compressed when the large mass comes to rest is given by the absolute value of x.
By following these steps, we can find the value of d.
To find the distance d by which the spring is compressed when the large mass comes briefly to rest, we need to analyze the scenario step by step.
Step 1: Find the initial momentum of the steel ball:
Using the equation for momentum, p = mv, where m is the mass and v is the velocity, the initial momentum of the steel ball is given by:
p_initial = m_initial * v_initial
Since the steel ball loses no kinetic energy during the collision, the velocity after the collision will be the same as the initial velocity but in the opposite direction, so the momentum after the collision is:
p_final = -m_initial * v_initial
Step 2: Find the initial kinetic energy of the steel ball:
The initial kinetic energy of the steel ball is given by the equation:
KE_initial = (1/2) * m_initial * v_initial^2
Step 3: Transfer of momentum:
When the steel ball collides with the large mass, momentum is transferred from the steel ball to the large mass. Since there is no loss of kinetic energy, the momentum transferred is equal to the change in momentum of the steel ball.
Step 4: Find the final velocity of the large mass:
Assuming the large mass moves to the left and comes to a brief stop before moving backward due to the compressed spring, we set up the equation for momentum:
m_large * v_final_large = - p_initial_steel_ball
where m_large is the mass of the large mass and v_final_large is its final velocity.
Step 5: Determine the force exerted by the frictional surface on the large mass:
The force exerted by the frictional surface opposes the motion, and the magnitude of the force is given by:
f_friction = μ * m_large * g
where μ is the kinetic coefficient of friction and g is the acceleration due to gravity.
Step 6: Determine the deceleration of the large mass:
The deceleration is given by Newton's second law of motion, which states that:
f_net = m_large * a
The net force acting on the large mass is the force due to the spring and the force of friction, so we have:
-k * d - f_friction = m_large * a
Substituting the value of f_friction from Step 5, we get:
-k * d - μ * m_large * g = m_large * a
Step 7: Find the final distance d by which the spring is compressed:
Assuming the large mass comes to rest after compressing the spring, the final velocity is 0 m/s. From Step 4, we know that:
v_final_large = 0
Substituting this into the equation from Step 6, we get:
-k * d - μ * m_large * g = 0
From this equation, we can solve for d, the distance by which the spring is compressed.