Two ice skaters approach each other at right
angles. Skater A has a mass of 30.7 kg and
travels in the +x direction at 2.64 m/s. Skater
B has a mass of 109 kg and is moving in the
+y direction at 1.29 m/s. They collide and
cling together.
Find the final speed of the couple.
Answer in units of m/s.
To solve this problem, 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 velocity. Let's denote the final speed of the couple as Vf.
Before the collision:
Skater A momentum (pA) = mass of Skater A * velocity of Skater A
pA = (30.7 kg) * (2.64 m/s) = 80.808 kg·m/s in the +x direction
Skater B momentum (pB) = mass of Skater B * velocity of Skater B
pB = (109 kg) * (1.29 m/s) = 140.61 kg·m/s in the +y direction
Since the skaters move at right angles, the x and y components of momentum are independent of each other and can be summed separately.
Initial x-component momentum = Final x-component momentum
pAx + pBx = (mA + mB) * Vfx
Since Skater A is moving entirely in the x-direction and Skater B is moving entirely in the y-direction, there is no x-component of momentum for skater B. Therefore, pBx = 0.
80.808 kg·m/s = (30.7 kg + 109 kg) * Vfx
80.808 kg·m/s = 139.7 kg * Vfx
Vfx = 80.808 kg·m/s / 139.7 kg = 0.578 m/s
The final speed of the couple is 0.578 m/s.
To find the final speed of the couple after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision should be equal to the total momentum after the collision.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v). Mathematically, momentum can be represented as:
p = m × v
Before the collision, Skater A has a momentum in the +x direction, and Skater B has a momentum in the +y direction. So we need to break their velocities into components.
Let's denote the velocity of Skater A as v_A and the velocity of Skater B as v_B. The initial velocities can be expressed as:
v_A = 2.64 m/s (in the +x direction)
v_B = 1.29 m/s (in the +y direction)
Now, we can calculate the initial momentum for each skater separately.
The momentum of Skater A (p_A) in the +x direction can be calculated as:
p_A = m_A × v_A
= 30.7 kg × 2.64 m/s
The momentum of Skater B (p_B) in the +y direction can be calculated as:
p_B = m_B × v_B
= 109 kg × 1.29 m/s
Since the two skaters collide and cling together, their final momentum, p_final, will be the sum of their initial momenta.
p_final = p_A + p_B
Since momentum is a vector quantity, we need to consider both the magnitude and direction for each momentum component. The final speed of the couple can be calculated using the Pythagorean theorem:
final speed = √((p_final_x)^2 + (p_final_y)^2)
Let's calculate the final speed step by step:
1. Calculate the initial momentum of Skater A:
p_A = 30.7 kg × 2.64 m/s
2. Calculate the initial momentum of Skater B:
p_B = 109 kg × 1.29 m/s
3. Calculate the resultant momentum in the x direction:
p_final_x = p_A
4. Calculate the resultant momentum in the y direction:
p_final_y = p_B
5. Calculate the final speed using the Pythagorean theorem:
final speed = √((p_final_x)^2 + (p_final_y)^2)
Ma*Va + Mb*Vb = Ma*V + Mb*V.
30.7*2.64 + 109*1.29i + 30.7V + 109V.
81.05 + 140.6i = 139.7V.
162.3m/s[60o] = 139.7V.
V = 1.16m/s[60o] N. of E.