Physics

A 3.0kg air puck sliding at 4.0m/s collides with a 5.0kg puck. After the collision the 3.0kg puck is moving north at 1.0m/s while the 5.0kg puck is moving south at 1.0m/s.
a) What was the initial velocity of the 5kg puck?

I did:
m1v1 + m2v2 = m1v1' + m2v2'
(3)(4) + (5v2) = (3)(1) + (5)(1)
and got -0.8m/s for v2.

b) What percentage of the initial kinetic energy remained after the collision?
I did
KE1' + KE2'
----------
KE1 + KE2
and got 16%

Now these questions I'm completely lost with because of the angles.

2) A 3kg ball moving east strikes a 2kg ball that is moving east at 2m/s. Following the collision the 3kg ball moves at 4m/s in a direction [E55S] while the other ball moves in a direction of [E35N].

a) By considering components of momentum in a N-S direction, calculate the speed that the 2kg ball had after the collision.
b) By considering conservation of momentum in an east-west direction, find the initial speed of the 3kg ball.

3) A 20kg child sitting on smooth ice is sliding straight north at 2m/s when she throws a big 2kg snow ball which (after the throw) moves [N60E] at 10m/s.
a) Calculate the momentum, speed and direction of motion of the child following the throw.

  1. 👍 0
  2. 👎 0
  3. 👁 393
  1. 2) A 3kg ball moving east strikes a 2kg ball that is moving east at 2m/s. Following the collision the 3kg ball moves at 4m/s in a direction [E55S] while the other ball moves in a direction of [E35N].

    a) By considering components of momentum in a N-S direction, calculate the speed that the 2kg ball had after the collision.
    -------------
    a)
    Initial momentum N is zero
    final momentum N:
    0 = 2 (s sin 35) - 3 (4 sin 55)
    s = 12 sin 55 / 2 sin 35 = 9.83/1.15
    s = 8.55 m/s
    =========================

    b) By considering conservation of momentum in an east-west direction, find the initial speed of the 3kg ball.
    --------------------------------
    initial momentum E = 3 v + 2*2 = 4 + 3 v
    final momentum E = 2*8.55 cos35 + 3*4 cos 55
    so
    4+3v = 14+6.88
    3 v = 16.88
    v = 5.63

    1. 👍 0
    2. 👎 0
  2. What about the answer to #3?

    1. 👍 0
    2. 👎 0
  3. 1a. Given:
    M1 = 3kg, V1 = 4 m/s.
    M2 = 5 kg, V2 = ?
    V3 = 1i m/s = Velocity of M1 after collision.
    V4 = -1i m/s = Velocity of M2 after collision.

    Momentum before = Momentum after.
    M1*V1 + M2*V2 = M1*V3 + M2*V4.
    3*4 + 5*V2 = 3i + (-5i),
    5V2 = -12 - 2i = 12.17[9.5o],
    V2 = 2.43m/s[9.5o] S. of W. = 2.43m/s[189.5o] CCW.

    b. KEb = 0.5M1*V1^2 + 0.5M2*V2 = 0.5*3*4^2 + 0.5*5*2.43^2 = 38.8 J. = Kinetic energy before collision.
    KEa = 0.5M1*V3^2 + 0.5M2*V4^2 = 0.5*3*1^2 + 0.5*5*(-1)^2 = 4 J. = KE after collision.
    4/38.8 * 100% = 10.3% remains.

    2.

    1. 👍 0
    2. 👎 0
  4. 2. Given:
    M1 = 3 kg, V1 = ?
    M2 = 2 kg, V2 = 2 m/s.
    V3 = 4 m/s[E55S] = 4m/s[145o]CW = Velocity of M1 after collision.
    V4 = ? [E35N] = ?[55o]CW = Velocity of M2 after collision.

    M1*V1 + M2*V2 = M1*V3 + M2*V4.
    3*V1 + 2*2 = 3*4[145] + 2*V4[55].
    3V1 + 4 = 12[145] + 2V4[55],
    Divide both sides by 2:
    Eq1: 1.5V1 + 2 = 6[145] + V4[55],

    a. V3 = (V1(M1-M2) + 2M2*V2)/(M1+ M2) = 4[145o].
    (V1(1) + 8)/(5) = 4[145],
    (V1 + 8)/5 = 4[145],
    V1 + 8 = 20[145],
    V1 = 20[145] - 8 = 11.5 + (-16.4)I - 8 = 3.5 - 16.4i = 16.8m/s[-12o].
    V1 = 16.8m/s[E12S] = 16.8m/s[102o]CW. = Velocity of M1 before collision.

    b. In Eq1, replace V1 with 16.8[102o] and solve for V4.
    1.5*16.8[102] + 2 = 6[145] + V4[55].
    24.65+(-5.24i) + 2 = 3.44+(-4.91i) + V4[55],
    26.65 - 5.24i = 3.44 - 4.91i + V4[55],
    23.21 - 0.33i = V4[55],
    0.82V4 + 0.57V4i = 23.21 - 0.33i,
    V4(0.82 + 0.57i) = 23.21[-89.2] = 23.21[179.2] CW,
    V4(1[55.2] = 23.21[179.2],
    V4 = 23.21[179.2] / 1[55.2] = 23.21m/s[124o] CW. = Velocity of M2 after collision.

    1. 👍 0
    2. 👎 0

Respond to this Question

First Name

Your Response

Similar Questions

  1. physics (this one is killing me clear answer)

    At time t = 0 s, a puck is sliding on a horizontal table with a velocity 3.60 m/s, 35.0° above the +x axis. As the puck slides, a constant acceleration acts on it that has the following components: ax = –0.360 m/s2 and ay =

  2. Physics

    At time t = 0s, a puck is sliding on a horizontal table with a velocity 3.00 m/s, 65.0 above the +x axis. As the puck slides, a constant acceleration acts on it that was the following components: ax = 0.460m/s2 and ay = 0.980m/s2.

  3. Physics

    A hockey puck is sliding across a frozen pond with an initial speed of 6.5 m/s. It comes to rest after sliding a distance of 7.6 m. What is the coefficient of kinetic friction between the puck and the ice? I know this one is not

  4. physics

    An air puck of mass m1 = 0.45 kg is tied to a string and allowed to revolve in a circle of radius R = 1.0 m on a frictionless horizontal table. The other end of the string passes through a hole in the center of the table, and a

  1. physic

    A hockey player strikes a puck that is initially at rest. The force exerted by the stick on the puck is 950 N, and the stick is in contact with the puck for 4.4 ms (0.0044 s). (a) Find the impulse imparted by the stick to the

  2. Physics

    On a frictionless, horizontal air table, puck A (with mass 0.25kg) is moving toward puck B (with mass 0.35kg), which is initially at rest.After the collision,puck A has a velocity of 0.12m/s to the left, and puck B has a velocity

  3. physics

    a 160g hockey puck slides horizontally along the ice, and the coefficient of kinetic friction between puck and ice is 0.15. if the puck starts sliding at 3 m/s before it comes to a stop it travels 3.6m how do you get to this

  4. PHYSICS

    In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 1.43 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck

  1. physics

    a hockey puck is sliding across frictionless ice at 6m/s. a)what is the net force on the puck? b)what forces act on the puck?

  2. physics

    A 0.400-kg ice puck, moving east with a speed of 5.38 m/s , has a head-on collision with a 0.950-kg puck initially at rest. Assume that the collision is perfectly elastic. What is the speed of the 0.400-kg puck after the

  3. Physics

    In attempting to pass the puck to a teammate, a hockey player gives it an initial speed of 2.95 m/s. However, this speed is inadequate to compensate for the kinetic friction between the puck and the ice. As a result, the puck

  4. Physics

    A hockey player hits a puck with his stick, giving the puck an initial speed of 5.0 m/s. If the puck slows uniformly and comes to rest in a distance of 20 m, what is the coefficient of kinetic friction between the ice and the

You can view more similar questions or ask a new question.