A 45 kg student runs down the sidewalk at a speed of 4.20m/s and jumps onto the stationary skateboard. The student and skateboard then move down the sidewalk at 3.95 m/s. What type of collision is this? What is the mass of the skateboard?
conservation of momentum
45*4.2+0=(45+M)3.95
solve for Mass M.
Now to see if it was an elastic collision (it wont be when the masses are joined in a collision), do this.
does final energy=initial energy ?
1/2 45 4.2^2=? 1/2 (45+M)3.95^2
check that.
2s
To determine the type of collision, we can apply the principle of conservation of momentum.
Let's denote the mass of the student as m1 and the mass of the skateboard as m2.
According to the conservation of momentum, the total momentum before the collision should be equal to the total momentum after the collision.
Initial momentum (before the collision) = Final momentum (after the collision)
Initially, only the student is moving, so the initial momentum is given by:
Initial momentum = m1 * v1, where v1 is the velocity of the student (4.20 m/s)
After the collision, both the student and the skateboard are moving together with a velocity of 3.95 m/s. Hence, the final momentum is given by:
Final momentum = (m1 + m2) * v2, where v2 is the velocity after the collision (3.95 m/s)
Since the momentum is conserved:
m1 * v1 = (m1 + m2) * v2
To determine the type of collision, we need to calculate the mass of the skateboard (m2).
Using the equation above, we can rearrange it to solve for m2:
m1 * v1 = (m1 + m2) * v2
(m1 * v1) / v2 = m1 + m2
m2 = (m1 * v1) / v2 - m1
Now, substitute the given values:
m1 = 45 kg
v1 = 4.20 m/s
v2 = 3.95 m/s
m2 = (45 kg * 4.20 m/s) / 3.95 m/s - 45 kg
After performing the calculation, the mass of the skateboard is found to be approximately 20.5 kg.
Now, to determine the type of collision, consider that in an elastic collision, both momentum and kinetic energy are conserved. In an inelastic collision, only momentum is conserved.
Since no information about kinetic energy is given, we can conclude that this collision is an inelastic collision.
To determine the type of collision and the mass of the skateboard, we can use the principles of conservation of momentum.
The momentum of an object is calculated by multiplying its mass by its velocity.
Initial momentum = final momentum
The initial momentum of the student running can be calculated as:
Initial momentum = mass of the student × velocity of the student
m1 × v1 = (45 kg) × (4.20 m/s)
The final momentum of the student and skateboard moving together can be calculated as:
Final momentum = (mass of the student + mass of the skateboard) × velocity of both
(m1 + m2) × v2 = (45 kg + m2) × (3.95 m/s)
Since the student jumps onto the skateboard, we can assume that there are no external forces acting on the system. Therefore, the collision is a perfectly elastic collision.
In a perfectly elastic collision, both momentum and kinetic energy are conserved.
To solve for the mass of the skateboard (m2), we need to set up an equation using the principles of conservation of momentum:
m1 × v1 = (m1 + m2) × v2
Using the given values:
(45 kg) × (4.20 m/s) = (45 kg + m2) × (3.95 m/s)
Now, we can solve for m2:
(45 kg) × (4.20 m/s) = (45 kg + m2) × (3.95 m/s)
189 kg·m/s = (45 kg + m2) × (3.95 m/s)
To solve for m2, we need to isolate it on one side of the equation. Let's start by dividing both sides of the equation by 3.95 m/s:
(189 kg·m/s) ÷ (3.95 m/s) = 45 kg + m2
48 kg = 45 kg + m2
Now, we can solve for m2 by subtracting 45 kg from both sides:
48 kg - 45 kg = m2
Therefore, the mass of the skateboard (m2) is 3 kg.
In summary, the collision between the student and skateboard is a perfectly elastic collision, and the mass of the skateboard is 3 kg.