In: Physics
1) A ball 1 with a mass of 2.0 kg and moving at 2.0 m/s strikes
a glancing blow on a second ball 2 which is initially at rest.
Assume no external forces act. After the collision, ball 1 is
moving at right angles to its original direction at a speed of 3.0
m/s.
(a) Calculate the initial momentum of the system.
(b) Determine the magnitude of the momentum of Ball 2 after the
collision?
(c) In what direction is Ball 2 moving after the collision?
(d) If Ball 2 has a mass of 5.0 kg, what is its speed after the
collision?
(e) If the impact lasted 0.020 s, calculate the magnitude of the
average force exerted on B during the collision.
(f) Based on the kinetic energy before and after the collision
calculate classify the collision as (inelastic, completely
inelastic, or fully elastic)
M1 = 2.0 Kg
(V1)i = 2m/s i ( let us fix this as our X-axis).
(V1)f = 3.0 m/s j
(V2)i = 0 m/s
a)
Initial Momentum of the system = 2.0 Kg x 2m/s + 0 = 4 Kg-m/s i
b)
(V2)f = ?
To conserve momentum along X-axis and also along Y-axis
4 kg-m/s i. = (M2 x (V2) x Cos ) i , where is angle made by M2 with respect to X-axis
Momentum of Ball 1 after collision = 3.0 m/s x 2.0 Kg = 6.0 Kg-m/s j
To conserve momentum along Y-axis
6.0 Kg-m/s j +(M2 x (V2)x sin ) (-j) = 0
So Component of momentum of ball 2 after collision along -Y-axis = 6.0 Kgm/s (-j)
Component of momentum of ball 2 after collision along X-axis = 4 Kg-m/s ( i )
Total momentum of ball 2 after collision = ( 4.0 i - 6.0 j) Kg-m/s
c)
direction of ball 2 after collision = Arc Tan (- 6.0/4.0) = -56.3 deg with respect to +ve X-axis
d)
M2 = 5 Kg
V2 = (1/5) ( 4.0 i -6.0 j) = 0.8 i - 1.2 j
Magnitude of V2 = sqrt( 0.8^2+1.2^2) = 1.44 m/s
e)
F . = (2) = (4.0 i - 6.0 j)/ 0.020 s = 200 i - 300 j
f)
KE before Collision = 0.5 x 2.0 Kg x (2.0 m/s)^2 = 4 J
KE after collision = 0.5 [ 2.0 kg x (3.0m/s)^2 + 5 kg x ( 1.44m/s)^2 ]= 14.2 J
So this collision is not possible as this violates conservation of energy . The Energy after collision can not be greater than Energy before collision.
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