1) A 1 kg mass moving at 1 m/s has a totally inelastic collision with a 0.7 kg mass. What is the speed of the resulting combined mass after the collision?
2) A cart of mass 1 kg moving at a speed of 0.5 m/s collides elastically with a cart of mass kg at rest. The speed of the second mass after the collision is 0.667 m/s. What is the speed 1 kg mass after the collision?
3) A 0.010 kg bullet is shot from a 0.500 kg gun at a speed of 230 m/s. Find the speed of the gun.
4)Two carts with a masses of 4 kg and 3 kg move toward each other on a frictionless track with speeds of 5.0 m/s and 4 kg m/s respectively. The carts stick together after the colliding head on. Find the final speed.
5) A cart of mass 1.5 kg moving at a speed of 1.2 m/s collides elastically with a cart of mass 1.0 kg moving at a speed of 0.75 m/s. (the carts are moving at the same direction)The speed
of the second mass (1.0 kg) after the collision is 0.85 m/s. What is the speed
of the 1.5 kg mass after the collision?
please help me!
>>FORMULA's<<
W=Fd
F=ma
W=mad
J=Ft or J=p
Ft=mvf-mvi
P=w/t
P=energy/time
P=mad/t
P= Fd/t
>> I'm so sorry, please help me! :) and Thank You!!
5. M1 = 1.5kg, V1 = 1.2 m/s.
M2 = 1.0kg, V2 = 0.75 m/s.
V3 = Velocity of M1 after the collision.
V4 = 0.85 m/s. = Velocity of M2 after the collision.
Conservation of KE Eq.
V3 = (V1(M1-M2) + 2M2*V2)/(M1+M2).
V3=(1.2(1.5-1.0)+2*0.75)/(1.5+1.0) = 0.84 m/s. = Velocity of M1 after the collision.
Correction:
5. M1*V1 + M2*V2 = M1*V3 + M2*V4.
1.5*1.2 + 1.0*0.75 = 1.5*V3 + 1.0*V4,
1.5*V3 + 1.0*V4 = 2.55,
Replace V4 with 0.85 and solve for V3:
1.5*V3 + 0.85 = 2.55,
1.5V3 = 2.55-0.85 = 1.7,
V3 = 1.13 m/s. = Velocity of M1 after the collision.
I'd be happy to help you with these questions step-by-step! Let's go through each one:
1) In a totally inelastic collision, the two masses stick together after the collision. To find the speed of the resulting combined mass, we can use the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and its velocity. So, before the collision, the momentum of the 1 kg mass is given by 1 kg * 1 m/s = 1 kg m/s. The momentum of the 0.7 kg mass is 0.7 kg * 0 m/s (as it is at rest) = 0 kg m/s.
Since the total momentum before the collision is equal to the total momentum after the collision, we can set up the equation: initial momentum = final momentum
1 kg m/s + 0 kg m/s = (1 kg + 0.7 kg) * vf, where vf is the final velocity of the combined mass.
Simplifying the equation, we get: 1 kg m/s = 1.7 kg * vf
Solving for vf, we find that vf = 0.588 m/s. Therefore, the speed of the resulting combined mass after the collision is approximately 0.588 m/s.
2) In an elastic collision, both momentum and kinetic energy are conserved. To find the speed of the 1 kg mass after the collision, we can use the law of conservation of momentum and kinetic energy.
Before the collision, the momentum of the 1 kg cart is given by 1 kg * 0.5 m/s = 0.5 kg m/s. The momentum of the second mass is 2 kg * 0 m/s (as it is at rest) = 0 kg m/s.
Since the total momentum before the collision is equal to the total momentum after the collision, we can set up the equation: 0.5 kg m/s = (1 kg + 2 kg) * vf1, where vf1 is the final velocity of the 1 kg cart.
Simplifying the equation, we get: 0.5 kg m/s = 3 kg * vf1
Solving for vf1, we find that vf1 = 0.167 m/s. Therefore, the speed of the 1 kg mass after the collision is approximately 0.167 m/s.
3) To find the speed of the gun after the bullet is shot, we can use the law of conservation of momentum.
Before the shot, the momentum of the bullet is given by 0.010 kg * 230 m/s = 2.3 kg m/s. The momentum of the gun is 0.500 kg * 0 m/s (as it is at rest) = 0 kg m/s.
Since the total momentum before the shot is equal to the total momentum after the shot, we can set up the equation: 2.3 kg m/s = (0.500 kg + mass of the gun) * vf2, where vf2 is the final velocity of the gun.
Simplifying the equation, we get: 2.3 kg m/s = (0.500 kg + mass of the gun) * vf2
We need the mass of the gun to calculate vf2. Could you provide that information?
4) In this case, since the carts stick together after the collision, momentum is conserved. We can use the law of conservation of momentum to find the final speed.
The total momentum before the collision is given by (4 kg * 5.0 m/s) + (3 kg * -4 m/s) = 15 kg m/s - 12 kg m/s = 3 kg m/s.
Since the total momentum before the collision is equal to the total momentum after the collision, we can set up the equation: 3 kg m/s = (4 kg + 3 kg) * vf3, where vf3 is the final velocity of the two carts stuck together.
Simplifying the equation, we get: 3 kg m/s = 7 kg * vf3
Solving for vf3, we find that vf3 = 0.429 m/s. Therefore, the final speed of the two carts stuck together is approximately 0.429 m/s.
5) In this case, since the collision is elastic, both momentum and kinetic energy are conserved. We can use the law of conservation of momentum and kinetic energy to find the speed of the 1.5 kg mass after the collision.
Before the collision, the momentum of the 1.5 kg cart is given by 1.5 kg * 1.2 m/s = 1.8 kg m/s. The momentum of the second mass is 1.0 kg * 0.75 m/s = 0.75 kg m/s.
Since the total momentum before the collision is equal to the total momentum after the collision, we can set up the equation: 1.8 kg m/s + 0.75 kg m/s = (1.5 kg + 1.0 kg) * vf4, where vf4 is the final velocity of the 1.5 kg cart.
Simplifying the equation, we get: 2.55 kg m/s = 2.5 kg * vf4
Solving for vf4, we find that vf4 = 1.02 m/s. Therefore, the speed of the 1.5 kg mass after the collision is approximately 1.02 m/s.
I hope these explanations help! Let me know if you have any further questions.
Sure, I can help you with these physics problems. Let's go through each question and explain the steps needed to solve them.
1) To find the speed of the resulting combined mass after an inelastic collision, we need to use the concept of conservation of momentum. In an inelastic collision, the two masses stick together and move with the same velocity after the collision.
The equation we can use is:
(m1 * v1 + m2 * v2) = (m1 + m2) * vf
Given that m1 = 1 kg, v1 = 1 m/s, m2 = 0.7 kg, and we want to find vf, we can plug in these values and solve for vf.
2) For an elastic collision, both momentum and kinetic energy are conserved. We can use the equations for conservation of momentum and kinetic energy to solve this problem.
The equation for momentum is the same as in the previous question:
(m1 * v1 + m2 * v2) = (m1 * vf1 + m2 * vf2)
And the equation for kinetic energy conservation is:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * vf1^2 + (1/2) * m2 * vf2^2
Given m1 = 1 kg, v1 = 0.5 m/s, v2 = 0 (since the second mass is at rest), vf2 = 0.667 m/s, we need to find vf1.
3) To find the speed of the gun after a bullet is shot, we can use the conservation of momentum. Before the shot, the gun and bullet are at rest, so the initial momentum is zero. After the shot, the momentum of the bullet and gun together should be zero as well.
The equation is:
m1 * v1 + m2 * v2 = 0
Given m1 = 0.01 kg, v1 = 230 m/s, m2 = 0.5 kg, we need to find v2.
4) To find the final speed after two carts collide and stick together, we again use the conservation of momentum.
The equation is:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
Given m1 = 4 kg, v1 = 5 m/s, m2 = 3 kg, v2 = -4 m/s (since the second cart is moving in the opposite direction), we need to find vf.
5) For an elastic collision of two carts moving in the same direction, we can use the same conservation of momentum and kinetic energy equations as in question 2.
The equations are:
(m1 * v1 + m2 * v2) = (m1 * vf1 + m2 * vf2)
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * vf1^2 + (1/2) * m2 * vf2^2
Given m1 = 1.5 kg, v1 = 1.2 m/s, m2 = 1.0 kg, v2 = 0.75 m/s, vf2 = 0.85 m/s, we need to find vf1.
Now that you understand the steps, you can plug in the given values into the respective equations and solve for the unknowns.