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'
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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.
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.
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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
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b) By considering conservation of momentum in an east-west direction, find the initial speed of the 3kg ball.
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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
What about the answer to #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.
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.
To solve these problems, we need to use the principles of conservation of momentum and apply them to the given scenarios.
1) For the first scenario, we are given the masses and velocities of two pucks before and after a collision. We need to find the initial velocity of the 5kg puck and the percentage of initial kinetic energy that remained after the collision.
a) To find the initial velocity of the 5kg puck, you correctly used the equation:
m1v1 + m2v2 = m1v1' + m2v2'
Plugging in the given values:
(3)(4) + (5)(v2) = (3)(1) + (5)(1)
Simplifying the equation:
12 + 5v2 = 3 + 5
5v2 = -8
v2 = -1.6 m/s
Therefore, the initial velocity of the 5kg puck was -1.6m/s (moving southwards).
b) To find the percentage of initial kinetic energy that remained after the collision, you correctly used the equation:
(KE1' + KE2') / (KE1 + KE2) * 100%
Plugging in the values:
KE1' = (1/2)(3)(1^2) = 1.5 J (Kinetic energy of the 3kg puck after the collision)
KE2' = (1/2)(5)(-1^2) = -2.5 J (Kinetic energy of the 5kg puck after the collision)
KE1 = (1/2)(3)(4^2) = 24 J (Initial kinetic energy of the 3kg puck)
KE2 = (1/2)(5)(-0.8^2) = 1 J (Initial kinetic energy of the 5kg puck)
Plugging in the values:
(1.5 - 2.5) / (24 + 1) * 100% = -1 / 25 * 100% = -4%
So, the percentage of initial kinetic energy that remained after the collision is -4%.
2) For the second scenario, we need to find the speed of the 2kg ball after the collision and the initial speed of the 3kg ball.
a) By considering components of momentum in a N-S direction, we can set up the equations:
For the 2kg ball:
2kg * v2 = 2kg * sin(35°) * v2' (Considering the N-S component)
Here, v2' is the speed of the 2kg ball after the collision.
Simplifying the equation:
v2 = sin(35°) * v2'
b) By considering conservation of momentum in the east-west direction, we can set up the equation:
For the 3kg ball:
3kg * 3m/s = 3kg * cos(55°) * v1
Here, v1 is the initial speed of the 3kg ball.
Simplifying the equation:
9 = cos(55°) * v1
Now, you can solve these two equations simultaneously to find the values of v2' and v1.
3) For the third scenario, we need to calculate the momentum, speed, and direction of motion of the child following the throw.
Given:
Mass of the child (m1) = 20kg
Initial velocity of the child (v1) = 2m/s
Mass of the snowball (m2) = 2kg
Final velocity of the snowball (v2') = 10m/s (N60E)
To calculate the momentum, we use the equation:
p = m1v1 + m2v2
Plugging in the values:
p = (20kg)(2m/s) + (2kg)(10m/s) = 40kg·m/s + 20kg·m/s = 60kg·m/s
To calculate the speed of the child, we use the equation:
speed = |v| = √(v_x^2 + v_y^2)
Given that the child is sliding straight north initially, the horizontal component of velocity remains 0 after the throw, and the vertical component can be calculated:
v_y = m1v1_y + m2v2'_y
v_y = (20kg)(0m/s) + (2kg)(10sin60°m/s) = 20√3 m/s
So, the speed of the child after the throw is:
speed = √(0^2 + (20√3)^2) = √(0 + 1200) = √1200 ≈ 34.64 m/s
Lastly, to find the direction of motion of the child, we can use trigonometry to find the angle between the north direction and the velocity vector of the child. Since the child is moving directly north, the angle would be 0°.
Therefore, the momentum of the child is 60kg·m/s, the speed is approximately 34.64 m/s (heading north), and the direction of motion is directly north.