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 are the speed and the direction of the resulting 50 g ball of clay?

Answer 1

The speed and the direction of the resulting 50 g ball of clay is 1.70 m/s at a direction 45° north of east.

Explanation:

Here momentum is conserved.

Initial momentum = Final momentum

Let east represent positive x axis and north represent positive y axis.

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

Mass of clay 1 = 20g = 0.02 kg

Velocity of clay 1 = 3 i m/s

Mass of clay 2 = 30g = 0.03 kg

Velocity of clay 3 = 2 j m/s

Initial momentum = 0.02 x 3 i + 0.03 x 2 j

Initial momentum = 0.06 i + 0.06 j

Combined mass of clay = 50 g = 0.05 kg

We need to find the velocity of combined clay.

Final momentum =  0.06 i + 0.06 j

0.05 x Velocity =  0.06 i + 0.06 j

Velocity = 1.2 i + 1.2 j

Magnitude $=\sqrt{1.2^{2} +1.2^2} =1.70 m/s$

Angle =$tan^{-1}(\frac{1.2}{1.2} )=45^0$

The speed and the direction of the resulting 50 g ball of clay is 1.70 m/s at a direction 45° north of east.

Answer 2

Answer:

1.697 m/s

Explanation:

m1 = 20 g

u1 = 3 m/s east = 3 i

m2 = 30 g

u2 = 2 m/s north = 2 j

Let the velocity is v after the collision

Use conservation of momentum

$m_{1}\overrightarrow{u_{1}}+m_{2}\overrightarrow{u_{2}}= \left ( m_{1}+m_{2} \right )\overrightarrow{v}$

$20\times 3\widehat{i}+30\times 2\widehat{j}= 50\overrightarrow{v}$

$1.2\widehat{i}+1.2\widehat{j}= \overrightarrow{v}$

v = 1.697 m/s