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1) A billiard ball rolling across a table at 1.25 m/s makes a head-on elastic co

ID: 1533020 • Letter: 1

Question

1) A billiard ball rolling across a table at 1.25 m/s makes a head-on elastic collision with an identical ball. Find the speed of each ball after the collision when each of the following occurs.

The second ball is moving toward the first at a speed of 1.15 m/s.

**1st ball??: ____m/s

2nd ball: 1.25 m/s

2) A 68.0 kg person throws a 0.0500 kg snowball forward with a ground speed of 33.0 m/s. A second person, with a mass of 59.0 kg, catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of 2.20m/s, and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard the friction between the skates and the ice.

thrower 2.2016 m/s **catcher??   
_____ m/s

Explanation / Answer


ELASTIC

m1 = m                             m2 = m


speeds before collision


u1 = 1.25 m/s                          u2 = -1.15 m/s

speeds after collision


v1 = ?                              v2 = ?


initial momentum before collision


Pi = m1*u1 + m2*u2

after collision final momentum

Pf = m1*v1 + m2*v2

from momentum conservation


total momentum is conserved

Pf = Pi


m1*u1 + m2*u2 = m1*v1 + m2*v2

m1*(u1-v1) = m2*(v2-u2) .....(1)


from energy conservation


total kinetic energy before collision = total kinetic energy after collision


KEi = 0.5*m1*u1^2 + 0.5*m2*u2^2


KEf =   0.5*m1*v1^2 + 0.5*m2*v2^2


KEi = KEf


0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

0.5*m1*(u1^2-v1^2) = 0.5*m2*(v2^2-u2^2).....(2)

solving 1&2


we get

u1 + v1 = v2 + u2


u1 - u2 = v2-v1


v2 = u1 - u2 + v1 ..........(3)

using 3 in 1

v1 = ((m1-m2)*u1 + (2*m2*u2))/(m1+m2)


v2 = ((m2-m1)*u2 + (2*m1*u1))/(m1+m2)

speed of first ball v1 = ((m-m)*1.25-(2*m*1.15))/(m+m) = -1.15 m/s

speed of second ball v2 = (-(m-m)*1.15+(2*m*1.25))/(m+m) = 1.25 m/s


=====================

2)

thrower

initial momentum pi = (68+0.05)*2.2


final momentum Pf = (68*v) + (0.05*33)


from momentum conservation


Pf = Pi

(68*v) + (0.05*33) = (68+0.05)*2.2


v = 2.1773 m/s <<<----------answer


catcher

initial momentum Pi = 59*0


final momentum Pf = (59+0.05)*vf

Pf = pi

vf = 0 <<<<----answer