A 68 kg man is ice-skating due north with a velocity of 5.6 m/s when he collides with a 42 kg child. The man and child stay together and have a velocity of 3.0 m/s at an angle of 35° north of due east immediately after the collision. What are (a) the magnitude and (b) the direction (as an angle measured from due East) of the child's velocity just before the collision?
To find the magnitude and direction of the child's velocity just before the collision, we can use the principle of conservation of linear momentum.
Step 1: Find the initial momentum of the man.
The momentum of an object is calculated by multiplying its mass by its velocity.
Momentum (man) = mass (man) × velocity (man)
= 68 kg × 5.6 m/s
= 380.8 kg·m/s
Step 2: Find the initial momentum of the child.
Momentum (child) = mass (child) × velocity (child)
We don't know the child's velocity just yet, so let's assume it is v m/s.
Momentum (child) = 42 kg × v
Step 3: Express the final momentum of the system after the collision.
Since the man and child stick together after the collision, their final momentum will be the sum of their individual momenta.
Final momentum (man and child) = (mass (man) + mass (child)) × velocity (man and child)
= (68 kg + 42 kg) × 3.0 m/s
= 110 kg × 3.0 m/s
= 330 kg·m/s
Step 4: Apply the law of conservation of linear momentum.
According to the conservation of linear momentum, the initial momentum of the system (before the collision) should be equal to the final momentum of the system (after the collision).
Initial momentum (man) + Initial momentum (child) = Final momentum (man and child)
380.8 kg·m/s + 42 kg × v = 330 kg·m/s
Step 5: Solve for the child's velocity just before the collision.
380.8 kg·m/s + 42 kg × v = 330 kg·m/s
42 kg × v = 330 kg·m/s - 380.8 kg·m/s
42 kg × v = -50.8 kg·m/s
v = -50.8 kg·m/s / 42 kg
v ≈ -1.21 m/s
The child's velocity just before the collision is approximately -1.21 m/s.
Step 6: Find the magnitude and direction of the child's velocity.
The magnitude of the child's velocity is the absolute value of the velocity, so the magnitude is 1.21 m/s.
To find the direction, we can use trigonometry. The child's velocity is at an angle of 35° north of due east. Since "due east" is a reference point, we can subtract 35° from 90° to find the direction from due east.
Direction from due east = 90° - 35°
= 55° east of due east
Therefore, the (a) magnitude and (b) direction of the child's velocity just before the collision are:
(a) The magnitude is 1.21 m/s.
(b) The direction is 55° east of due east.
To find the magnitude and direction of the child's velocity just before the collision, we can use the principle of conservation of momentum.
Considering the collision as an isolated system, the total momentum before the collision must be equal to the total momentum after the collision.
The momentum of an object is given by the product of its mass and velocity:
Momentum = mass × velocity
Let's assume the child's velocity just before the collision is v, and the man's velocity just before the collision is u.
The given information can be summarized as follows:
Mass of the man (m₁) = 68 kg
Velocity of the man just before the collision (u₁) = 5.6 m/s
Mass of the child (m₂) = 42 kg
Velocity of the child just before the collision (v₂) = v (unknown)
After the collision:
Velocity of the combined system (man and child) = 3.0 m/s
Angle of the combined velocity (θ) = 35° north of due east
Knowing that momentum is a vector quantity, we can write the momentum equations in two components: one in the north direction and the other in the east direction.
The momentum conservation equations are:
Momentum before collision in the north direction:
m₁u₁ + m₂v₂ = (m₁ + m₂)vcos(θ)
Momentum before collision in the east direction:
0 = (m₁ + m₂)vsin(θ)
Let's solve these equations step by step:
First, let's find the magnitude of the child's velocity just before the collision (v).
Using the momentum equation in the east direction, we can see that 0 = (m₁ + m₂)vsin(θ), which means that sin(θ) = 0, as sin(0) = 0.
Therefore, we can conclude that the angle θ = 0°, which means the child's velocity just before the collision was along the east direction, i.e., in the positive x-axis.
Now let's substitute this information into the momentum equation in the north direction:
m₁u₁ + m₂v₂ = (m₁ + m₂)vcos(θ)
Plugging in the known values:
(68 kg)(5.6 m/s) + (42 kg)(v₂) = (68 kg + 42 kg)(3.0 m/s)(cos 0°)
Simplifying the equation:
380.8 + 42v₂ = 110(3.0)
380.8 + 42v₂ = 330
Rearranging the equation:
42v₂ = 330 - 380.8
42v₂ = -50.8
Dividing by 42:
v₂ = -1.2095 m/s
Since the magnitude of a velocity cannot be negative, we take the absolute value:
|v₂| = 1.2095 m/s
Therefore, the magnitude of the child's velocity just before the collision is approximately 1.2095 m/s.
The direction of the child's velocity just before the collision, as an angle measured from due East, is 0°.
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