A 0.145 kg block is released from rest from
the top of a 51.7◦ frictionless incline. When
it has fallen a vertical distance of 1.3 m, a
0.00794 kg bullet is fired into the block along
a path parallel to the slope of the incline, and
momentarily brings the block to rest, stopping
in the block.
The acceleration of gravity is 9.8 m/s2 .
a) Find the speed of the bullet just before
impact.
Answer in units of m/s.
b) What bullet speed is needed to send the block
up the incline to its initial position?
Answer in units of m/s.
for part a:
v= (mass of block/mass of bullet)(sqrt2gh)
v = (0.145/0.00794) x (sqrt 2x9.8x1.3)
v = 18.262 x 5.0478
v= do the calculation
I did not notice your part b. That is a harder question.
Use energy conservation to predict the velocity V' (after impact) needed to bring the block and bullet together back to the top of the incline.
(1/2)(M+m)V'^2 = (M+m)g*1.4
V' = 5.05 m/s (same as V in part a, but in the opposite drection)
Now apply conversation of momentum to the impact event, and require that V' (UP the incline) be the final velocity.
-M V + m v = (M +m) V
v = (2M/m +1) V
a) Well, let's tackle the first part, shall we? To find the speed of the bullet just before impact, we need to apply some physics shenanigans.
First, we know that the block is brought to rest by the bullet. This means that the final velocity of the block, after the bullet has hit it, is 0 m/s.
Using the principle of conservation of momentum, we know that the initial momentum of the bullet and the block must be zero as well (since the bullet comes to a stop in the block).
The momentum of an object is given by the product of its mass and velocity. Since the block is at rest initially, its momentum is 0 kg m/s.
So, if we let v be the speed of the bullet just before impact, and m be the mass of the bullet, we have:
momentum of bullet = momentum of block
m * v = 0
Solving for v, we find that v = 0 m/s.
Wait, what? That can't be right! Looks like I made a miscalculation. Let me try again!
(a) The block and the bullet have the same final velocity (0 m/s) after the collision. Applying the law of conservation of momentum, we have:
(m_b * v_b) + (m_bullet * v_bullet) = 0
Where m_b is the mass of the block and v_b is its final velocity, and m_bullet is the mass of the bullet and v_bullet is the velocity of the bullet before impact.
We know the mass of the block (m_b = 0.145 kg) and the mass of the bullet (m_bullet = 0.00794 kg). We need to find v_bullet.
Since the block is released from rest on a frictionless incline, we can use the equations of motion to determine the final velocity of the block (v_b) after it has fallen a vertical distance of 1.3 m.
Using the equation:
v_b^2 = u^2 + 2as
where u is the initial velocity (0 m/s), s is the distance (1.3 m), and a is the acceleration (due to gravity, -9.8 m/s^2), we get:
v_b^2 = 0 + 2 * (-9.8) * 1.3
v_b^2 = -25.48
v_b ≈ -5.048 m/s
Since velocity is a vector quantity, the negative sign indicates that the velocity is in the opposite direction of the downward motion. We can ignore the negative sign for now and plug v_b into our momentum equation:
(m_b * v_b) + (m_bullet * v_bullet) = 0
(0.145 * -5.048) + (0.00794 * v_bullet) = 0
-0.73076 + (0.00794 * v_bullet) = 0
0.00794 * v_bullet = 0.73076
v_bullet ≈ 91.8 m/s
So, the speed of the bullet just before impact is approximately 91.8 m/s.
b) Now, let's move on to the second part. We want to find the bullet speed needed to send the block up the incline to its initial position.
In this case, the final velocity of the block (v_b) is not zero since it has to move up the incline and reach its initial position. We need to find the bullet speed (v_bullet) that will give us the desired final velocity.
Using the same momentum conservation equation as before (m_b * v_b) + (m_bullet * v_bullet) = 0, we can rearrange it to solve for v_bullet:
v_bullet = -(m_b * v_b) / m_bullet
v_bullet = -(0.145 * v_b) / 0.00794
Substituting the value we found for v_b (-5.048 m/s):
v_bullet = -(0.145 * -5.048) / 0.00794
v_bullet ≈ 92.36 m/s
So, the bullet speed needed to send the block up the incline to its initial position is approximately 92.36 m/s.
Oh, look! I've got my numbers all mixed up again! My sincerest apologies for that. Let me try one more time!
(b) To send the block up the incline to its initial position, the final velocity of the block (v_b) should be equal to its initial velocity (0 m/s). Using the conservation of momentum equation:
(m_b * v_b) + (m_bullet * v_bullet) = 0
where m_b is the mass of the block, v_b is its final velocity, m_bullet is the mass of the bullet, and v_bullet is the velocity of the bullet before impact.
Since v_b is 0 m/s and m_b is 0.145 kg, we can solve for v_bullet:
(0.145 * 0) + (0.00794 * v_bullet) = 0
0 + 0.00794 * v_bullet = 0
0.00794 * v_bullet = 0
v_bullet = 0 / 0.00794
v_bullet = 0 m/s
Well, that's no fun! It seems like something went wrong again. My apologies for the confusion. It looks like I need to double-check my calculations one more time.
To solve this problem, we can use the principle of conservation of energy.
a) First, let's calculate the potential energy of the block just before the bullet is fired.
The formula for potential energy is given by:
PE = m * g * h
where m is the mass of the block, g is the acceleration due to gravity, and h is the vertical distance.
PE = (0.145 kg) * (9.8 m/s^2) * (1.3 m)
PE = 1.846 J
Since the incline is frictionless, the only other form of energy the block has is kinetic energy just before the bullet is fired.
Next, let's calculate the kinetic energy of the block just before the bullet is fired.
The formula for kinetic energy is given by:
KE = (1/2) * m * v^2
where m is the mass of the block, and v is the velocity/speed of the block.
KE = (1/2) * (0.145 kg) * v^2
According to the conservation of energy principle, the total mechanical energy before the bullet is fired is equal to the total mechanical energy after the bullet is fired.
Total Mechanical Energy Before = Total Mechanical Energy After
So, we can set up an equation as follows:
PE + KE (before) = KE (after)
1.846 J + (1/2) * (0.145 kg) * v^2 = 0 (since the block is momentarily brought to rest)
Simplifying the equation:
v^2 = - (2 * 1.846 J) / (0.145 kg)
v^2 = - 25.393 J/kg
Since velocity cannot be negative, we take the positive square root:
v = √(25.393 J/kg)
v ≈ 5.04 m/s
Therefore, the speed of the bullet just before impact is approximately 5.04 m/s.
b) To find the bullet speed needed to send the block up the incline to its initial position, we need to consider the change in potential energy.
This change in potential energy is equal to the work done by the bullet on the block.
The work done (W) is given by:
W = Force * Distance * cos(θ)
Since the incline is frictionless, the only force acting on the block is the force of gravity.
The distance is the vertical distance the block needs to travel.
W = m * g * h * cos(θ)
In this case, θ is 51.7 degrees, so we need to convert it to radians:
θ_rad = 51.7° * (π/180°)
Then, we can substitute the values into the equation:
W = (0.145 kg) * (9.8 m/s^2) * (1.3 m) * cos(θ_rad)
W ≈ 19.399 J
The work done by the bullet is equal to the change in potential energy:
W = Change in PE
19.399 J = m * g * h
Now, we can solve for the bullet speed (v) using the kinetic energy formula:
KE = (1/2) * m * v^2
19.399 J = (1/2) * (0.00794 kg) * v^2
Simplifying the equation:
v^2 = (19.399 J * 2) / (0.00794 kg)
v^2 = 4877.127 J/kg
Taking the positive square root:
v = √(4877.127 J/kg)
v ≈ 69.90 m/s
Therefore, the bullet speed needed to send the block up the incline to its initial position is approximately 69.90 m/s.
First calculate the speed V of the sliding block before it is hit by the bullet. You can use conservation of energy for that, since the incline is frictionless.
(1/2) M V^2 = M g * 1.3 m
V = sqrt(2*9.8*1.3) = 5.05 m/s
The bullet must have an equal and opposite momentum before impact, since the momentum right after impact is zero.
m v = M V
where m is the bullet mass and v is the bullet velocity.
Solve for v.
AFTER impact, gravity will cause the bullet and block to resume falling together, but you are only worried about the instant after impact, before gravity provides an impulse to increase the momentum again.