Question

A 20 g ball of clay traveling east at 3.0 m/s collides with a 30 g ball of clay traveling north at 2.0 m/s. What is the magnitude and direction of initial momentum for each ball before collision? a. What is the momentum of 50 g blob of clay in x? What is the speed in that same direction? b. What is the momentum of 50 g blob of clay in y? What is the speed in that same direction? c. What is the resultant speed of the blob? d. What is direction of the blob?

Answer 1

Part A.

Momentum is given by:

P = mass*Velocity = m*V

m1 = mass of ball 1 = 20 gm = 0.02 kg

m2 = mass of ball 2 = 30 gm = 0.03 kg

u1 = Initial velocity of m1 = 3.0 m/sec towards east

So, u1_x = 3.0 m/sec & u1_y = 0 m/sec

u2 = Initial velocity of m2 = 2.0 m/sec towards North

So, u2_x = 0 m/sec & u2_y = 2.0 m/sec

Now Initial momentum of each ball before collision will be:

P1 = m1*u1 = 0.02*3.0 = 0.06 kg-m/sec

Magnitude of P1 = |P1| = 0.06 kg-m/sec, Direction = towards east

P2 = m2*u2 = 0.03*2.0 = 0.06 kg-m/sec

Magnitude of P2 = |P2| = 0.06 kg-m/sec, Direction = towards North

Part B.

Now Using momentum conservation in x-direction:

Pi_x = Pf_x

m1*u1_x + m2*u2_x = Pf_x

Pf_x = 0.02*3.0 + 0.03*0 = 0.06 kg-m/sec

Momentum of 50 gm blob in x-direction = 0.06 kg-m/sec

Now Since

Pf_x = (m1 + m2)*V_x

V_x = Speed of 50 g blob in x-direction = Pf_x/(m1 + m2)

V_x = 0.06/(0.03 + 0.02) = 1.2 m/sec towards east

Part C.

Now Using momentum conservation in y-direction:

Pi_y = Pf_y

m1*u1_y + m2*u2_y = Pf_y

Pf_y = 0.02*0 + 0.03*2.0 = 0.06 kg-m/sec

Momentum of 50 gm blob in y-direction = 0.06 kg-m/sec

Now Since

Pf_y = (m1 + m2)*V_y

V_y = Speed of 50 g blob in y-direction = Pf_y/(m1 + m2)

V_y = 0.06/(0.03 + 0.02) = 1.2 m/sec towards North

Part D.

Resultant velocity of blob will be:

V = V_x i + V_y j

V = (1.2 i + 1.2 j) m/sec

So speed of blob will be:

|V| = sqrt (1.2^2 + 1.2^2)

|V| = 1.7 m/sec

Direction of blob = arctan(V_y/V_x)

Direction = arctan(1.2/1.2) = arctan (1) = 45 deg from +ve x-axis

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