Question

A 3.4 g particle moving at 18 m/s collides with a 0.82 g particle initially at rest. After the collision the two particles have velocities that are directed at equal angles of 51 ◦ on either side of the original line of motion of the 3.4 g particle. What is the speed of the 0.82 g particle after the collision?

Answer 1

Answer:

4.34 m/s

Explanation:

m1 = 3.4 g

m2 = 0.82 g

u1 = 18 m/s

u2 = 0

θ = 51°

Let the velocities after the collision be v1 and v2 respectively.

Use conservation of momentum along X axis

m1 x u1 + m2 x u2 = m1 x v1 x Cos 51 + m2 x v2 x Cos 51

3.4 x 18 + 0 = 3.4 x v1 x 0.74 + 0.82 x v2 x 0.74

61.2 = 2.5 v1 + 0.61 v2     ... (1)

Use conservation of momentum along Y axis

0 = m1 x v1 x Sin 51 - m2 x v2 x Sin 51

3.4 x v1 x 0.67 = 18 x v2 x 0.67

2.23 x v1 = 12.06 v2

v1 = 5.41 v2

Substitute in equation (1)

61.2 = 2.5 x 5.41 v2 + 0.61 v2

61.2 = 14.1 v2

v2 = 4.34 m/s

Answer 2

The speed of the 0.82g particle after the collision is; 4.34m/s.

Inelastic Collision

From the task content;

• It follows that, the velocities after the collision are v1 and v2 respectively.

It therefore follows from conservation of momentum along X axis that;

• (m1 x u1) + (m2 x u2) = (m1 x v1 x Cos 51) + (m2 x v2 x Cos 51)

• (3.4 x 18) + 0 = (3.4 x 0.74) x v1 + (0.82 x 0.74) x v1

• 61.2 = 2.5v1 + 0.61v2.......equation (1)

Similarly, it follows from conservation of momentum along Y axis that;

0 = m1 x v1 x Sin 51 - m2 x v2 x Sin 51

3.4 x v1 x 0.67 = 18 x v2 x 0.67

2.23 x v1 = 12.06 v2

v1 = 5.41 v2

Upon Substitution into equation (1), it follows that

• 61.2 = 2.5 x 5.41 v2 + 0.61 v2

• 61.2 = 14.1 v2

v2 = 4.34 m/s

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