Two bodies, one of mass 4kg and the other of mass 6 kg, approach each other head-on with speed of 2m/s and 8m/s, respectively. Calculate the velocity of each body after collision if the collision is perfectly elastic.
Given: M1 = 4kg, V1 = -2m/s.
M2 = 6kg, V2 = 8m/s.
V3 = velocity of M1 after collision.
V4 = velocity of M2 after collision.
Momentum before = Momentum after
M1*V1 + M2*V2 = M1*V3 + M2*V4
4*(-2) + 6*8 = 4*V3 + 6*V4
Eq1: 4V3 + 6V4 = 40.
V3 = ((M1-M2)V1 + 2M2*V2)/(M1+M2)
V3 = ((4-6)(-2) + 12*8)/(4+6) = (4 + 96)/10 = 10 m/s.
In Eq1, replace V3 with 10 and solve for V4:
4*10 + 6V4 = 40
V4 = 0.
equestrian worker accidentally drop a brick from a high scaffold what is the velocity of a break after 4.0s
To solve this problem, we can use the conservation of momentum and kinetic energy in an elastic collision.
The conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it can be expressed as:
m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final
Where:
m1 and m2 are the masses of the two bodies
v1_initial and v2_initial are the initial velocities of the two bodies
v1_final and v2_final are the final velocities of the two bodies
Now, let's calculate the velocities of each body after the collision:
Step 1: Substitute the given values into the equation.
(4 kg * 2 m/s) + (6 kg * 8 m/s) = 4 kg * v1_final + 6 kg * v2_final
Step 2: Simplify the equation.
(8 kg * m/s) + (48 kg * m/s) = 4 kg * v1_final + 6 kg * v2_final
Step 3: Rearrange the equation to solve for v1_final.
4 kg * v1_final = (8 kg * m/s) + (48 kg * m/s) - 6 kg * v2_final
v1_final = [(8 kg * m/s) + (48 kg * m/s) - 6 kg * v2_final] / 4 kg
Step 4: Substitute the values back into the equation to solve for v1_final.
v1_final = (8 kg * m/s) + (48 kg * m/s) - 6 kg * v2_final) / 4 kg
Step 5: Solve for v2_final by substituting the value of v1_final into the equation.
v2_final = [4 kg * v1_final - (8 kg * m/s) - (48 kg * m/s)] / 6 kg
Step 6: Substitute the value of v1_final into the equation to solve for v2_final.
v2_final = [4 kg * [(8 kg * m/s) + (48 kg * m/s) - 6 kg * v2_final] / 4 kg] - (8 kg * m/s) - (48 kg * m/s)] / 6 kg
Simplifying this equation further will give us the values of v1_final and v2_final.
To calculate the velocity of each body after a perfectly elastic collision, we can use the principle of conservation of momentum and kinetic energy.
The principle of conservation of momentum states that the total momentum before the collision is equal to the total momentum after the collision. Mathematically, it can be written as:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
Where:
m1 and m2 are the masses of the bodies
v1 and v2 are the initial velocities of the bodies
v1' and v2' are the final velocities of the bodies
In this case, let's denote the body with mass 4kg as body 1 and the body with mass 6kg as body 2.
Given:
m1 = 4 kg (mass of body 1)
m2 = 6 kg (mass of body 2)
v1 = 2 m/s (initial velocity of body 1)
v2 = 8 m/s (initial velocity of body 2)
For a perfectly elastic collision, the kinetic energy of the system is conserved. Mathematically, it can be written as:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
Now we have two equations – one for the conservation of momentum and one for the conservation of kinetic energy. Solving these equations simultaneously will give us the final velocities of the bodies after the collision.
Let's start by rearranging the conservation of momentum equation:
m1 * v1 + m2 * v2 = m1 * v1' + m2 * v2'
(4 kg * 2 m/s) + (6 kg * 8 m/s) = 4 kg * v1' + 6 kg * v2'
8 kg m/s + 48 kg m/s = 4 kg * v1' + 6 kg * v2'
56 kg m/s = 4 kg * v1' + 6 kg * v2' ---(Equation 1)
Now, let's rearrange the conservation of kinetic energy equation:
(1/2) * m1 * v1^2 + (1/2) * m2 * v2^2 = (1/2) * m1 * v1'^2 + (1/2) * m2 * v2'^2
(1/2) * 4 kg * (2 m/s)^2 + (1/2) * 6 kg * (8 m/s)^2 = (1/2) * 4 kg * v1'^2 + (1/2) * 6 kg * v2'^2
(1/2) * 4 kg * 4 m^2/s^2 + (1/2) * 6 kg * 64 m^2/s^2 = (1/2) * 4 kg * v1'^2 + (1/2) * 6 kg * v2'^2
8 kg m^2/s^2 + 192 kg m^2/s^2 = 4 kg * v1'^2 + 6 kg * v2'^2 ---(Equation 2)
Now, we have two equations (Equation 1 and Equation 2) with two unknowns v1' and v2'. We can solve these equations simultaneously to find the values of v1' and v2'.
Next, substitute Equation 1 into Equation 2:
56 kg m/s = 4 kg * v1' + 6 kg * v2'
8 kg m^2/s^2 + 192 kg m^2/s^2 = 4 kg * v1'^2 + 6 kg * v2'^2
Now we have a system of two equations:
4 * v1' + 6 * v2' = 56
4 * v1'^2 + 6 * v2'^2 = 200
Solving this system of equations will give us the values of v1' and v2', which are the final velocities of the bodies after the collision.