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

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

Answer 2

The final velocity of the two balls after collision is 1.7 m/s and the direction is 45⁰.

The given parameters;

• mass of the first clay ball, m₁ = 20 g = 0.02 kg
• initial velocity of the first clay ball, u₁ = 3 m/s
• mass of the second clay ball, m₂ = 30 g = 0.03 kg
• initial velocity of the second ball, u₂ = 2 m/s

The initial momentum of the fist ball is calculated as follows;

P₁ = m₁u₁

P₁ = (0.02)(3)

P₁ = 0.06 kg.m/s

The initial momentum of the second ball is calculated as follows

P₂ = m₂u₂

P₂ = (0.03)(2)

P₂ = 0.06 kg.m/s

The resultant initial momentum of the two balls is calculated as follows;

$P_i = \sqrt{P_x_i ^2 + P_y_i^2} \\\\P_i = \sqrt{0.06^2 + 0.06^2} \\\\P_i = 0.085 \ kg.m/s$

Apply the principle of conservation of momentum, to determine the final velocity of the two balls;

$P_f = P_i\\\\mv= 0.085\\\\v = \frac{0.085}{0.05} \\\\v = 1.7 \ m/s$

The direction of the two ball's velocity is calculated as follows;

$\theta = tan^{-1}(\frac{P_y_i}{P_x_i} )\\\\\theta = tan^{-1}(\frac{0.06}{0.06} )\\\\\theta = 45 \ ^0$

Thus, the final velocity of the two balls after collision is 1.7 m/s and the direction is 45⁰.

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