1- Consider two masses, m p and m t , on a surface where friction is negligible.

Question

1- Consider two masses, mp and mt, on a surface where friction is

negligible. Assume that mt is initially at rest and that mass mp,

having a velocity v = vpi i, collides with it. The diagram to the

right shows the arrangement. After the collision, mt (the target)

and mp (the projectile), travel at angles ? and ?, respectively.

Write the expression for conservation of momentum in the xand

y-directions

2- . One way of doing experiments involving collisions in two

dimensions is to use an air hockey table. The air cushion between

the hockey pucks and table minimizes friction and conservation of

momentum applies; i. e., no horizontal force acts during the collision.

A second method is to collide the objects above the floor and allow

each object to be a projectile. With air resistance neglected, no

horizontal force acts on the objects during freefall, and each object

falls to the floor under the vertical force of gravity. If the collision

occurs a height H above the floor, show that the time t for each

object to hit the floor is t = (2H/g)1/2, where g is the acceleration due

to gravity.

3- If the target ball is not present, show that the initial momentum of the projectile po is given by

po = mp(Ro/t), where Ro is the range of the projectile when no collision occurs. The important

point to note here is that the range of the projectile is directly proportional to the projectile

Explanation / Answer

1) This is direct vector vector addition and conservation principle

in X diretion

initially mp *vpi
finally mp*vpf*cos(phi)+mt*vtf*cos(theta)

mp *vpi = mp*vpf*cos(phi)+mt*vtf*cos(theta)

in Y direction
initiall Zero
Finally
mt*vtf*sin(theta) - mp*vpf*sin(phi)

0 = mt*vtf*sin(theta) - mp*vpf*sin(phi)

2) H=ut+1/2*g*t^2
u=o no vertical speed after collision of balls (simple logic :P)
H= 1/2*g*t^2
t = (2H/g)1/2 hwnce proved!

3)

in third part too R=ut+1/2*a*t^2
unfortunately we dont have acceleration in horizontal direction which implies a=0

therefore R0=u*t

and p0=mp*u

hence from above po = mp(Ro/t)

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