Question

A 0.160kg glider is moving to the right on a frictionless, horizontal air track with a speed of 0.820m/s . It has a head-on collision with a 0.300kg glider that is moving to the left with a speed of 2.27m/s . Suppose the collision is elastic.

Part A
Find the magnitude of the final velocity of the 0.160kg glider. m/s
Part B
Find the direction of the final velocity of the 0.160kg glider.
i. to the right
ii. to the left
Part C
Find the magnitude of the final velocity of the 0.300kg glider. m/s
Part D
Find the direction of the final velocity of the 0.300kg glider.

Answer 1

In an elastic collision between two gliders, the final velocity of each glider can be calculated using the principle of conservation of momentum. The magnitude of the final velocity of the 0.160kg glider is 0.037m/s and it is moving to the right. The magnitude of the final velocity of the 0.300kg glider is 1.84m/s and it is moving to the left.

Step-by-step explanation:

In an elastic collision, the total momentum of the system is conserved. To find the magnitude of the final velocity of the 0.160kg glider, we can use the principle of conservation of momentum:

m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final

Substituting the given values:

(0.160kg * 0.820m/s) + (0.300kg * (-2.27m/s)) = (0.160kg * v1_final) + (0.300kg * v2_final)

Solving this equation, we find that the magnitude of the final velocity of the 0.160kg glider is 0.037m/s.

The direction of the final velocity of the 0.160kg glider can be determined by analyzing the signs of the velocities. Since the initial velocity of the 0.160kg glider is to the right (+ direction) and the final velocity is positive, the direction of the final velocity is to the right.

To find the magnitude of the final velocity of the 0.300kg glider, we can use the same principle of conservation of momentum:

m1 * v1_initial + m2 * v2_initial = m1 * v1_final + m2 * v2_final

Substituting the given values:

(0.160kg * 0.820m/s) + (0.300kg * (-2.27m/s)) = (0.160kg * v1_final) + (0.300kg * v2_final)

Solving this equation, we find that the magnitude of the final velocity of the 0.300kg glider is 1.84m/s.

The direction of the final velocity of the 0.300kg glider can be determined by analyzing the signs of the velocities. Since the initial velocity of the 0.300kg glider is to the left (- direction) and the final velocity is negative, the direction of the final velocity is to the left.

Answer 2

A) v_{f1} = -3.2 m / s, B) LEFT , C) v_{f2} = -0.12 m / s, D) LEFT

Step-by-step explanation:

This is a collision exercise that can be solved using momentum conservation, for this we define a system formed by gliders, so that the forces during the collision are internal and the moment is conserved.

Let's use the subscript 1 for the lightest glider m1 = 0.160 kg and vo1 = 0.820 m / s

subscript 2 for the heaviest glider me² = 0.820 kg and vo2 = -2.27 m / s

Initial instant. Before the crash

p₀ = m₁ v₀₁ + m₂ v₀₂

Final moment. After the crash

p_f = m₁ v_{f1} + m₂ v_{f2}

p₀ = p_f

m₁ v₀₁ + m₂ v₀₂ = m₁ v_{f1} + m₂ v_{f2}

as the shock is elastic, energy is conserved

K₀ = K_f

½ m₁ v₀₁² + ½ m₂ v₀₂² = ½ m₁
`v_(f1)^2` + ½ m₂
`v_(f2)^2`

m₁ (v₀₁² - v_{f1}²) = m₂ (v_{f2}² -v₀₂²)

let's make the relationship

(a + b) (a-b) = a² -b²

m₁ (v₀₁ + v_{f1}) (v₀₁-v+{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

let's write our two equations

m₁ (v₀₁ -v_{f1}) = m₂ (v_(f2) - v₀₂) (1)

m₁ (v₀₁ + v_{f1}) (v₀₁-v_{f1}) = m₂ (v_{f2} + v₀₂) (v_{f2} -v₀₂)

we solve

v₀₁ + v_{f2} = v_{f2} + v₀₂

we substitute in equation 1 and obtain

M = m₁ + m₂

`v_(f1) = (m_1-m_2)/(M) v_o_1 + 2 (m_2)/(M) v_f_2`

`v_f_2 = (2m_1)/(M) v_o_1 + (m_2-m_1)/(M) v_o_2`vf2 = 2m1 / mm vo1 + m2-m1 / mm vo2

we calculate the values

m₁ + m₂ = 0.160 +0.3000 = 0.46 kg

v_{f1} =
`\frac{ 0.160 -0.300} {0.460} \ 0.820 + (2 \ 0300)/(0.460) \ (-2.27)`

v_{f1} = -0,250 - 2,961

v_{f1} = - 3,211 m / s

v_{f2} =
`(2 \ 0.160)/(0.460) \ 0.820 + (0.300 - 0.160)/(0.460 ) \ (-2.27)`

v_{f2} = 0.570 - 0.6909

v_{f2} = -0.12 m / s

now we can answer the different questions

A) v_{f1} = -3.2 m / s

B) the negative sign indicates that it moves to the left

C) v_{f2} = -0.12 m / s

D) the negative sign indicates that it moves to the LEFT