two different masses have equal, non zero kinetic energies. the momentum of the smaller mass is?
what is the magnitude of momentum of a 0.140 baseball traveling at 45.0m/s?
a 2.00g bullet hits and becomes embedded in 5.00 kg wood block which is hanging from a 1.60m long string. this causes the block to swing through an arc of 5.50(degrees). what was the speed of the bulletbefore it hit the block ?
a 14000kg boxcar is coasting at 1.50m/d along a horizontal track when it hits and couples with a stationary 10000kg boxcar. what is the speed of the cars after the collision?
Let m be the smaller mass and v be the velocity of that mass. Let M be the larger mass and V be its velocity.
Because the energies are equal,
m v^2 = M V^2
(mv)^2/m = (MV)^2/M
mv/(MV) = sqrt (m/M) < 1
(since m/M < 1)
Therefore the smaller mass also has less momentum.
========================
You left out the dimensions of the mass of the baseball. I assume it is kg
Momentum magnitude is mass x velocity. You should know that. Do the calculation. Don't forget the units, which should be kg m/s
Your last two questions require you to apply the law of conservation of momentum
To answer these questions, we need to use the concepts of kinetic energy and momentum.
1. For two different masses with equal, non-zero kinetic energies:
The kinetic energy of an object is given by KE = (1/2)mv^2, where m is the mass of the object and v is its velocity. Since both masses have equal kinetic energies, their kinetic energy equations can be set equal to each other:
(1/2)m1v1^2 = (1/2)m2v2^2
Dividing both sides by (1/2), we get:
m1v1^2 = m2v2^2
2. To find the magnitude of momentum of a baseball:
Momentum is given by the equation p = mv, where p is the momentum, m is the mass, and v is the velocity. We are given the mass of the baseball (0.140 kg) and its velocity (45.0 m/s). We can plug these values into the equation to find the magnitude of the momentum.
3. To find the speed of the bullet before it hit the block:
In this case, we can apply the law of conservation of linear momentum. Before the collision, the bullet has momentum, and after the collision, the combined system of the block and bullet will have the same momentum. We can write this as:
m_bullet * v_bullet = (m_bullet + m_block) * v_combined
We are given the mass of the bullet (2.00 g = 0.00200 kg), the mass of the block (5.00 kg), and the swing arc of the block (5.50 degrees). The velocity of the combined system can be calculated from the swing arc using the equation v_combined = r * ω, where r is the length of the string (1.60 m) and ω is the angular velocity. To find ω, we need to convert the swing arc from degrees to radians by multiplying it with π/180. With all the values known, we can solve for v_bullet.
4. To find the speed of the cars after the collision:
Again, we can apply the law of conservation of momentum. Before the collision, the first boxcar has momentum, and after the collision, the combined system of the two boxcars will have the same momentum. We can write this as:
m1 * v1 = (m1 + m2) * v_combined
We are given the mass of the first boxcar (14,000 kg), the mass of the second boxcar (10,000 kg), and the speed of the first boxcar (1.50 m/s). We can solve for v_combined to find the speed of the cars after the collision.
By plugging in the given values and solving the equations, we can find the answers to these questions.
To answer these questions step-by-step:
1. For two different masses with equal, non-zero kinetic energies, we can use the formula for kinetic energy: KE = (1/2)mv^2, where KE is the kinetic energy, m is the mass, and v is the velocity.
Since the kinetic energies are equal, we can set up an equation:
(1/2)m1v1^2 = (1/2)m2v2^2
Since the masses are different, but the kinetic energies are equal, we can say that the momentum of the smaller mass is greater. So, the momentum of the smaller mass is greater.
2. To find the magnitude of momentum, we can use the formula: p = mv, where p is momentum, m is mass, and v is velocity.
Given:
Mass (m) = 0.140 kg
Velocity (v) = 45.0 m/s
Substituting the values into the formula:
p = (0.140 kg) * (45.0 m/s)
p = 6.3 kg m/s
So, the magnitude of momentum of the 0.140 kg baseball traveling at 45.0 m/s is 6.3 kg m/s.
3. To find the speed of the bullet before it hit the block, we need to apply the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Given:
Bullet mass (m1) = 2.00 g = 0.00200 kg
Wood block mass (m2) = 5.00 kg
Initial velocity of bullet (v1) = unknown
Final velocity of bullet-wood block system (v2) = 0 (since the bullet becomes embedded)
Using the conservation of momentum equation:
m1v1 = (m1 + m2)v2
Substituting the values:
(0.00200 kg)(v1) = (0.00200 kg + 5.00 kg)(0)
Simplifying the equation, since (0.00200 kg + 5.00 kg)(0) = 0:
(0.00200 kg)(v1) = 0
Therefore, the speed of the bullet before it hit the block is 0 m/s.
4. To find the speed of the cars after the collision, we can use the principle of conservation of momentum.
Given:
Mass of the first boxcar (m1) = 14000 kg
Initial velocity of the first boxcar (v1) = 1.50 m/s
Mass of the second boxcar (m2) = 10000 kg
Initial velocity of the second boxcar (v2) = 0 m/s (stationary)
Using the conservation of momentum equation:
(m1)(v1) + (m2)(v2) = (m1 + m2)(v)
Substituting the values:
(14000 kg)(1.50 m/s) + (10000 kg)(0 m/s) = (14000 kg + 10000 kg)(v)
21000 kg m/s = 24000 kg v
Dividing both sides by 24000 kg to solve for v:
v = (21000 kg m/s) / (24000 kg)
v ≈ 0.875 m/s
Therefore, the speed of the cars after the collision is approximately 0.875 m/s.