In a physics lab, 0.30 kg puck A, moving at 5.0 m/s [W], undergoes a collision with 0.40 kg puck B, which is initially at rest. Puck A moves off at 4.2 m/s [W 30° N]. Find the final velocity of puck B.
conserve momentum.
0.30<-5,0> = 0.30<-3.63,2.1> + 0.40<vx,vy>
solve for vx,vy the components of puck B's velocity
Given: M1 = 0.30 kg, V1 = -5.0 m/s.
M2 = 0.40 kg, V2 = 0.
V3 = 4.2 m/s[150o] CCW = velocity of M1 after collision.
V4 = velocity of M2 after collision.
M1*V1+M2*V2 = M1*V3+M2*V4.
0.30*(-5)+0.40*0 = 0.30*4.2[150o]+0.40*V4
-1.50 = 1.26[150o]+0.40*V4
-1.5 = -1.09+0.63i + 0.4*V4
-0.4*V4 = 0.41+0.63i
V4 = -1.025-1.575i = 1.88m/s[57o] S. of W. = velocity of puck B.
To find the final velocity of puck B after the collision, we can apply the principles of conservation of momentum. In an isolated system, the total momentum before a collision is equal to the total momentum after the collision.
Let's break down the given information:
Mass of puck A (m1): 0.30 kg
Initial velocity of puck A (v1i): 5.0 m/s [W]
Final velocity of puck A (v1f): 4.2 m/s [W 30° N]
Mass of puck B (m2): 0.40 kg
Initial velocity of puck B (v2i): 0 m/s
The total momentum before the collision is the sum of the individual momenta of pucks A and B:
Initial momentum (p_initial) = (m1 * v1i) + (m2 * v2i)
After the collision, the total momentum is the sum of the individual momenta of pucks A and B again:
Final momentum (p_final) = (m1 * v1f) + (m2 * v2f)
According to the law of conservation of momentum, p_initial = p_final.
Let's substitute the given values into the equation:
(m1 * v1i) + (m2 * v2i) = (m1 * v1f) + (m2 * v2f)
(0.30 kg * 5.0 m/s) + (0.40 kg * 0 m/s) = (0.30 kg * 4.2 m/s) + (0.40 kg * v2f)
Now, let's solve for v2f, the final velocity of puck B:
1.5 kg⋅m/s = 1.26 kg⋅m/s + (0.40 kg * v2f)
1.5 kg⋅m/s - 1.26 kg⋅m/s = 0.40 kg * v2f
0.24 kg⋅m/s = 0.40 kg * v2f
v2f = (0.24 kg⋅m/s) / 0.40 kg
v2f = 0.6 m/s
Therefore, the final velocity of puck B is 0.6 m/s.