A hockey puck moving at 0.4600 m/s collides with another puck that was at rest. The pucks have equal mass. The first puck is deflected 37.00° to the right and moves off at 0.3700 m/s. Find the speed and direction of the second puck after the collision.

.342m/s at 43 degrees to the right

initial x momentum = .46 m

initial y momentum = 0

final x momentum = .37 m cos 37 + v m cos T
final y momentum = .37 m sin 37 - v m sin T

so
.46 = .37 cos37 + v cos T
and
.37 sin 37 = v sin T

Given:

M1 = M kg, V1 = 0.46 m/s.
M2 = M kg, V2 = 0.
V3 = 0.37m/s[37o] = Velocity of M1 after collision.
V4 = ? = Velocity of M2 after collision.

Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V3 + + M2*V4.
M*0.46 + M*0 = M*0.37[37] + M*V4,
Divide both sides by M:
0.46 + 0 = 0.37[37o] + V4,
0.46 = 0.295+0.223i + V4,
V4 = 0.165 -.223i = 0.277[-53.5] = 53.5o S. of E. = 0.277m/s[306.5o] CCW.

Ahockey

To find the speed and direction 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 will be equal to the total momentum after the collision.

Let's consider the x and y components of momentum separately. The x component of momentum is given by:

Initial x momentum = (mass of first puck) * (speed of first puck)
Final x momentum = (mass of first puck) * (speed of first puck, after deflection)

Using the given values:
Initial x momentum = (mass) * (0.4600 m/s)
Final x momentum = (mass) * (0.3700 m/s)

We can rewrite these expressions in terms of the mass and speeds of both pucks, as they have equal mass:

Initial x momentum = (mass) * (0.4600 m/s)
Final x momentum = (mass) * (V2x)

The y component of momentum is also conserved, which means that:

Initial y momentum = 0 (since the second puck is initially at rest)
Final y momentum = 0 (since only the first puck is deflected horizontally)

Now, let's consider the momentum components in each direction. The x velocity of the second puck can be found using the formula for conservation of momentum in the x direction:

(mass) * (0.4600 m/s) = (mass) * (0.3700 m/s) + (mass of second puck) * (V2x)

Simplifying this equation, we get:

0.4600 m/s = 0.3700 m/s + V2x

V2x = 0.4600 m/s - 0.3700 m/s
V2x = 0.0900 m/s

So, the x component of the velocity of the second puck after the collision is 0.0900 m/s to the right. Since the second puck was initially at rest, its y component of velocity remains 0 after the collision.

Hence, the speed of the second puck after the collision is given by the magnitude of the final velocity vector:

V2 = sqrt((V2x)^2 + (V2y)^2)
V2 = sqrt((0.0900 m/s)^2 + (0 m/s)^2)
V2 = sqrt(0.0081 m^2/s^2)
V2 ≈ 0.0900 m/s

Therefore, the velocity (speed and direction) of the second puck after the collision is approximately 0.0900 m/s to the right.