Question

During the last shot in a curling match, you attempt a take-out to win the game. The two curling rocks are of equal mass, one red and the other blue, and collide in a glancing collision. The blue rock is initially at rest and is struck by the red rock moving initially East at 4.60 m/s. After the collision, the red rock moves in a direction that makes an angle of 38.0° North of East, while the blue rock makes an angle of 52.0° South of East. Determine the speed of each rock after the collision.

Answer 1

We know that after the collision, Momentum will remain conserved, So

Using Momentum conservation in x-direction:

Pix = Pfx

m1*u1x + m2*u2x = m1*v1x + m2*v2x

m1 = mass of red rock = m

m2 = mass of blue rock = m

u1x = Initial velocity of m1 in x-direction = 4.60 m/sec

u2x = Initial velocity of m2 in x-direction = 0 m/sec

v1x = Final velocity of m1 in x-direction = v1*cos 38 deg

v2x = Final velocity of m2 in x-direction = v2*cos 52 deg

So,

m*4.60 + m*0 = m*v1*cos 38 deg + m*v2*cos 52 deg

4.60 = v1*0.788 + v2*0.616

Now Using Momentum conservation in y-direction:

Piy = Pfy

m1*u1y + m2*u2y = m1*v1y + m2*v2y

m1 = mass of red rock = m

m2 = mass of blue rock = m

u1y = Initial velocity of m1 in y-direction = 0 m/sec

u2y = Initial velocity of m2 in y-direction = 0 m/sec

v1y = Final velocity of m1 in y-direction = v1*sin 38 deg

v2y = Final velocity of m2 in y-direction = -v2*sin 52 deg

So,

m*0 + m*0 = m*v1*sin 38 deg - m*v2*sin 52 deg

0 = v1*0.616 - v2*0.788

v1 = v2*(0.788/0.616) = v2*1.279

Now Using both Bold equation

4.60 = v1*0.788 + v2*0.616

v1 = v2*1.279

Substitute value of v1 in 1st equation:

4.60 = v2*1.279*0.788 + v2*0.616

v2 = 4.60/[1.279*0.788 + 0.616]

v2 = 2.833 m/sec = final speed of blue rock

Using this value

v1 = 2.833*1.279

v1 = 3.623 m/sec = final speed of red rock

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