A hockey puck of mass 0.16 kg, sliding on a nearly frictionless surface of ice with a velocity of 2.0 m/s (E), strikes a second puck at rest with a mass of 0.17 kg. the first puck has a velocity of 1.5 (N 31 E) after the collision. Determine the velocity of the second puck after the collision.
a proton flying to the east near the surface of the earth at 1000mls. what is the magnitude of the force the proton feels due to earths magnetic field
X: m₁v₁=m₁u₁cosα+m₂u₂cosβ
Y: 0=m₁u₁sinα - m₂u₂cosβ
m₁v₁-m₁u₁cosα = m₂u₂cosβ …(1)
m₁u₁sinα= m₂u₂cosβ ………...(2)
Divide (2) by (1)
tanβ= m₁u₁sinα/ (m₁v₁-m₁u₁cosα) =
=sinα/ (1- cosα) =3.6
β=74.5º
u₂=m₁u₁sinα/ m₂cosβ= 0.16•2•sin31/0.17•sin 74.5 =1 m/s
to Anton:
F=qvB? where q=e
v: m mile/s -> m/s
B= 5•10⁻⁵ T
4.5
Well, this sounds like quite the icy situation! Let me crunch some numbers for you and see what I come up with.
To solve this problem, we can use the principle of conservation of momentum. The total momentum before the collision should be equal to the total momentum after the collision.
Now, let's calculate the momentum of each puck before the collision. The momentum of an object is equal to its mass times its velocity. So, for the first puck:
Momentum of the first puck before the collision = mass of the first puck × velocity of the first puck
= (0.16 kg) × (2.0 m/s)
= 0.32 kg·m/s
The second puck is at rest initially, so its momentum before the collision is zero.
Next, let's calculate the total momentum after the collision. Since the first puck's velocity after the collision is given (1.5 m/s at N 31 E), we need to break it down into its horizontal and vertical components. Using some trigonometry, we find:
Velocity of the first puck (horizontal component) = 1.5 m/s × cos(31°)
= 1.5 m/s × 0.857 (rounded)
= 1.286 m/s
Velocity of the first puck (vertical component) = 1.5 m/s × sin(31°)
= 1.5 m/s × 0.530 (rounded)
= 0.795 m/s
Now, let's get back to calculating the total momentum after the collision. The total momentum after the collision should be equal to the sum of the momenta of both pucks. Since the second puck is initially at rest and its momentum is zero, we only need to consider the momentum of the first puck after the collision.
Total momentum after the collision = momentum of the first puck after the collision
= mass of the first puck × velocity of the first puck (after collision)
= (0.16 kg) × (velocity of the first puck (after collision))
= (0.16 kg) × (1.5 m/s at N 31 E)
= (0.16 kg) × (1.286 m/s) + (0.16 kg) × (0.795 m/s)
= 0.20576 kg·m/s + 0.1272 kg·m/s
= 0.33396 kg·m/s
Now, we know that the total momentum after the collision should be equal to zero for the second puck. So, we can set up an equation:
Total momentum after the collision = 0
= (mass of the second puck) × (velocity of the second puck (after collision))
Solving for the velocity of the second puck (after collision):
velocity of the second puck (after collision) = Total momentum after the collision / mass of the second puck
= 0 / (0.17 kg)
= 0 m/s
Therefore, the velocity of the second puck after the collision is 0 m/s. It doesn't budge! It appears that our second puck took a break during the collision and decided to stay at rest. Maybe it wanted to enjoy the ice a little longer or catch up on some Netflix. Who knows?
To determine the velocity of the second puck after the collision, we can use the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Momentum (p) is defined as the product of mass (m) and velocity (v):
p = m * v
Let's denote the initial velocity of the second puck as v2i (at rest) and the final velocity of the second puck as v2f (to be determined).
The momentum of the first puck before the collision can be calculated as:
p1i = m1 * v1i = 0.16 kg * 2.0 m/s = 0.32 kg·m/s (E)
The momentum of the second puck before the collision is zero because it is at rest:
p2i = m2 * v2i = 0.17 kg * 0 m/s = 0 kg·m/s
The momentum of the first puck after the collision can also be calculated as:
p1f = m1 * v1f = 0.16 kg * 1.5 m/s = 0.24 kg·m/s (N 31 E)
According to the principle of conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:
p1i + p2i = p1f + p2f
0.32 kg·m/s + 0 kg·m/s = 0.24 kg·m/s + p2f
0.32 kg·m/s = 0.24 kg·m/s + p2f
To isolate p2f, we can rearrange the equation:
p2f = 0.32 kg·m/s - 0.24 kg·m/s
p2f = 0.08 kg·m/s
Now, let's calculate the velocity of the second puck after the collision:
v2f = p2f / m2
v2f = 0.08 kg·m/s / 0.17 kg
v2f ≈ 0.47 m/s
Therefore, the velocity of the second puck after the collision is approximately 0.47 m/s.