two balls of equal mass,moving with speeds of 5 meter/seconds,collide head on.find the velocity of each after impact if (a)they are stuck together (b)the collision of perfectly elastic. (c)the coefficient of restitution in 1/3?
(a) 0
(b) 5 m/s and -5 m/s
(in opposite directions)
(c) sqrt(25/3) = 2.89 m/s
and -2.89 m/s
(in opposite directions)
a car accelerated from rest at 2 m/ s reached of 20 m/ s . it moved at this velocity for 50s and then started brake and stopped after 5s . find the total distance by this car from the starting point?
To find the velocities of the balls after the collision, we can use the principle of conservation of linear momentum. The formula for linear momentum is:
Momentum = mass × velocity
The total momentum before the collision is equal to the total momentum after the collision.
Given:
Mass of each ball = m
Initial velocity of the balls = 5 m/s
(a) When the balls are stuck together:
After the collision, the two balls stick together and move with a single velocity.
Let V be the velocity of the combined mass after the collision.
Initial momentum = (mass of ball 1 × velocity of ball 1) + (mass of ball 2 × velocity of ball 2)
= (m × 5) + (m × 5)
= 10m
Final momentum = (mass of combined mass × velocity of combined mass)
= (2m × V)
= 2mV
According to the conservation of linear momentum:
Initial momentum = Final momentum
10m = 2mV
Divide both sides by 2m to solve for V:
V = 10m / 2m
V = 5 m/s
So, when the balls are stuck together, the velocity after the impact is 5 m/s.
(b) When the collision is perfectly elastic:
In a perfectly elastic collision, both momentum and kinetic energy are conserved.
Let V1 and V2 be the velocities of the balls after impact.
The conservation of linear momentum can be written as:
(mass of ball 1 × velocity of ball 1) + (mass of ball 2 × velocity of ball 2) = (mass of ball 1 × velocity of ball 1 after impact) + (mass of ball 2 × velocity of ball 2 after impact)
Applying this equation:
(m × 5) + (m × 5) = (m × V1) + (m × V2)
Also, the conservation of kinetic energy can be written as:
(½ × mass of ball 1 × (velocity of ball 1)^2) + (½ × mass of ball 2 × (velocity of ball 2)^2) = (½ × mass of ball 1 × (velocity of ball 1 after impact)^2) + (½ × mass of ball 2 × (velocity of ball 2 after impact)^2)
Applying this equation:
(½ × m × (5)^2) + (½ × m × (5)^2) = (½ × m × (V1)^2) + (½ × m × (V2)^2)
Simplifying both equations, we have:
10m = mV1 + mV2 (1)
50m = m(V1^2) + m(V2^2) (2)
From equation (1), we can write:
V2 = 10m - V1
Substituting V2 in equation (2), we get:
50m = m(V1^2) + (10m - V1)^2
Expanding and simplifying:
50m = m(V1^2) + 100m - 20mV1 + (V1^2)
50m = 2m(V1^2) - 20mV1 + 100m
Divide the equation by m:
50 = 2V1^2 - 20V1 + 100
Rearrange the equation:
2V1^2 - 20V1 + 50 = 0
Solving this quadratic equation using the quadratic formula:
V1 = (-(-20) ± √((-20)^2 - 4×2×50)) / (2×2)
V1 = (20 ± √(400 - 400)) / 4
V1 = 20 / 4
V1 = 5 m/s
Using V2 = 10m - V1:
V2 = 10m - 5
V2 = 5 m/s
So, when the collision is perfectly elastic, the velocities of the balls after the impact are 5 m/s and 5 m/s.
(c) When the coefficient of restitution is 1/3:
The coefficient of restitution (e) is defined as the ratio of relative velocity after impact to the relative velocity before impact.
Let V1' and V2' be the velocities of the balls after impact.
The relative velocity before impact is the difference between the velocities of the balls:
Relative velocity before impact = velocity of ball 1 - velocity of ball 2
= 5 - 5
= 0
The relative velocity after impact is the difference between the velocities after the impact:
Relative velocity after impact = velocity of ball 1' - velocity of ball 2'
The coefficient of restitution can be expressed as:
e = (relative velocity after impact) / (relative velocity before impact)
Substituting the values:
1/3 = (V1' - V2') / 0
When the relative velocity before impact is 0, the coefficient of restitution cannot be determined. Therefore, we cannot determine the velocities of the balls after the impact when the coefficient of restitution is 1/3.
To solve this problem, we can apply the laws of conservation of momentum and kinetic energy to find the velocities of the balls after the collision.
(a) When the two balls are stuck together after the collision, the total mass of the system remains the same. Therefore, the total momentum before the collision is equal to the total momentum after the collision.
Step 1: Calculate the total momentum before the collision:
Initial momentum = Final momentum
m_1 * v_1_initial + m_2 * v_2_initial = (m_1 + m_2) * v_final
Since the masses (m_1 and m_2) are equal and the initial velocities (v_1_initial and v_2_initial) are the same:
2 * m * 5 m/s = 2 * m * v_final
Step 2: Solve for the final velocity (v_final):
v_final = 5 m/s
Therefore, the velocity of the balls after the collision, when they are stuck together, is 5 m/s.
(b) In a perfectly elastic collision, both momentum and kinetic energy are conserved.
Step 1: Calculate the total momentum before the collision:
Initial momentum = Final momentum
m * v_1_initial + m * v_2_initial = m * v_1_final + m * v_2_final
Since the masses (m_1 and m_2) are equal and the initial velocities (v_1_initial and v_2_initial) are the same:
2 * m * 5 m/s = m * v_1_final + m * v_2_final
Step 2: Calculate the total kinetic energy before and after the collision:
Initial kinetic energy = Final kinetic energy
(1/2) * m * (v_1_initial^2 + v_2_initial^2) = (1/2) * m * (v_1_final^2 + v_2_final^2)
Since the masses (m_1 and m_2) are equal and the initial velocities (v_1_initial and v_2_initial) are the same:
(1/2) * m * (2 * 5^2) = (1/2) * m * (v_1_final^2 + v_2_final^2)
50 = v_1_final^2 + v_2_final^2
Step 3: Solve the system of equations to find the final velocities:
From the momentum equation: v_1_final + v_2_final = 5 m/s
From the kinetic energy equation: v_1_final^2 + v_2_final^2 = 50
By solving these equations, we find v_1_final = 3 m/s and v_2_final = 2 m/s.
Therefore, the velocities of the balls after a perfectly elastic collision are v_1_final = 3 m/s and v_2_final = 2 m/s.
(c) When the coefficient of restitution (e) is given, we can use the formula:
e = (v_2_final - v_1_final) / (v_1_initial - v_2_initial)
Substituting the given values:
1/3 = (v_2_final - v_1_final)/(5 - (-5))
1/3 = (v_2_final - v_1_final)/10
Multiplying both sides by 10:
10/3 = v_2_final - v_1_final
Given that the masses and initial velocities are the same, we can assume that v_2_final = -2 v_1_final:
10/3 = -3 v_1_final
Dividing both sides by -3:
v_1_final = -10/9 m/s
Substituting this value in the equation v_2_final = -2 v_1_final:
v_2_final = -2 * (-10/9) m/s = 20/9 m/s (approximately 2.22 m/s)
Therefore, the velocities of the balls after the collision with a coefficient of restitution of 1/3 are v_1_final = -10/9 m/s (approximately -1.11 m/s) and v_2_final = 20/9 m/s (approximately 2.22 m/s).