Question

A 30.0 gg ball of clay traveling east at 1.00 m/sm/s collides and sticks together with a 40.0 gg ball of clay traveling north at 6.00 m/sm/s .

What is the speed of the resulting ball of clay?

Answer 1

m1 = 30 gm = 0.03 kg

m2 = 40 gm = 0.04 kg

taking east as X-direction, and norht as Y-direction

then

velocity of 1st clay =U1= 1.00 m/sec

velocity of 2nd clay = U2 = 6.00 m/sec

Resolivng the components along x and y direction,

U1x = 1.00 , U1y = 0.0

U2x = 0.00, U2y = 6.00

After the collision, the mass of resultant clay = M = 70 gm = 0.07 Kg

let velocity of the resultant clay = V

and lets its components along x and y direction are Vx and Vy

now applying lwa of conservation of linear momentum in X and Y-directions, we have following equations:

and

Putting the values in above equations:

we get

m/sec

and

m/sec

the resultant velocity (speed) =

Please check the units and modify if necessary