# 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.

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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

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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.

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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.

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