1.. Three collisions: a) Totally Inelastic collision: A 10-grams bullet is shot

Question

1.. Three collisions: a) Totally Inelastic collision: A 10-grams bullet is shot at a 2.0-kg wooden block and remains embedded in the block. Initially, the bullet and the block were moving towards each other as shown in the sketch below. Find the velocity of the block- bullet system right after the impact and calculate the fraction of the initial kinetic energy that Is lost during collision. Partially Inelastic collision; A 20-grams bullet Is shot at a 400-grams hard stone, Initially, the bullet and the block were moving towards each other as shown in the sketch below (the collision takes place above the ground). it is observed that right after the collision, the bullet is deflected upwards, perpendicular to its initial direction of motion, and with a third of its initial speed. Find the stone's speed and direction of motion right after the collision, and calculate the fraction of the initial kinetic energy that is lost during collision. c) Elastic collision: Two rubber balls (one of 200 grams mass and the other 300 grams) collide elastically head-on. Before the collision, they were moving as shown in the sketch below. Find the speed and direction of motion for each ball right after collision. (0 pts)

Explanation / Answer

1. (a) given that :

mass of the bullet, m1 = 10g = 0.01 kg

mass of the wooden block, m2 = 2 kg

initial speed of the bullet, v1 = 200 m/s

initial speed of the wooden block, v2 = 15 m/s

using an inelastic collisions,

m1 v1 + m2 v2 = (m1 + m2) V                                    { eq. 1 }

inserting the values in above eq.

(0.01 kg) (200 m/s) + (2 kg) (15 m/s) = (0.01 kg + 2 kg) V

(32 kg.m/s) = (2.01 kg) V

V = 15.9 m/s

the initial kinetic energy of the bullet & wooden block is given as :

K.Einitial = 1/2 m1 v12 + 1/2 m2 v22                             { eq. 2 }

inserting the values in eq.2,

K.Einitial = (0.5) [(0.01 kg) (200 m/s)2 + (2 kg) (15 m/s)2]

K.Einitial = (0.5) [(400 + 450)] J

K.Einitial = 425 J

the final kinetic energy of the system is given as :

K.Efinal = 1/2 [(m1 + m2) V2 ] { eq.3 }

inserting the values in eq.3,

K.Efinal = (0.5) [(0.01 kg + 2 kg) (15.9 m/s)2 ]

K.Efinal = (0.5) [(508.1)] J

K.Efinal = 254 J

fraction of the initial energy that is lost durina a collision which given as ::

f = K.Ei - K.Ef / K.Ei                             { eq. 4 }

inserting the values in eq.4,

f = [(425 J) - (254 J)] / (425 J)

f = 0.40

(c) applying the conservation of linear momentum,

m1 v1i2 + m2 v2i2 = m1 v1 + m2 v2                               { eq.5 }

the speed and direction of motion for each ball right after collision which is given as :

for an elastic collision, v1 = v1i (m1 - m2) / (m1 + m2) + v2i (2 m2) / (m1 + m2)                 { eq.6 }

where, m1 = mass of the first rubber ball = 200g = 0.2 kg

m2 = mass of the second rubber ball = 300g = 0.3 kg

v1i = initial speed of the first rubber ball = 22 m/s

v2i = initial speed of the second rubber ball = 10 m/s

inserting all these values in eq.6,

v1 = [(22 m/s) (0.2 kg - 0.3 kg) / (0.2 kg + 0.3 kg)] + [(10 m/s) (2 x 0.3 kg) / (0.2 kg + 0.3 kg)]

v1 = [(-2.2 / 0.5) + (6 / 0.5)] m/s

v1 = (3.8 / 0.5) m/s

v1 = 7.6 m/s

and
v2 = v1i (2 m1) / (m1 + m2) + v2i (m2 - m1) / (m2 + m1)                          { eq. 7 }

inserting the values in eq.7,

v2 = [(22 m/s) (2 x 0.2 kg) / (0.2 kg + 0.3 kg)] + [(10 m/s) (0.3 kg - 0.2 kg) / (0.3 kg + 0.2 kg)]

v2 = [(8.8 / 0.5) + (1 / 0.5)] m/s

v2 = (9.8 / 0.5) m/s

v2 = 19.6 m/s

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