Question

A 20 g ball of clay traveling east at 3.0 m/s collides with and sticks to a 30 g ball of clay traveling north at 2.0 m/s. What are the speed and direction of the resulting 50 g blob of clay?

Answer 1

Approximately
`1.7\; {\rm m\cdot s^(-1)}` north-east (
`45^(\circ)` east of north.)

Step-by-step explanation:

To find the speed and direction of the resulting object, start by finding its velocity (as a vector) using the conservation of momentum. The direction of velocity can be found by taking the arctangent of the ratio between the two components.

Momentum is equal to the product of mass and velocity. Because velocity is a vector quantity, momentum is also a vector quantity.

Let the first component denote motions in the north-south direction, with motions to the north being positive. Similarly, let the second component denote motions in the east-west direction, with motions to the east being positive.

Momentum should be conserved. In other words, the sum of momentum should stay the same before and after the collision. Before the collision, sum of momentum is:

`\displaystyle (20)\, \begin{bmatrix}0 \\ 3.0\end{bmatrix} + \displaystyle (30)\, \begin{bmatrix}2.0 \\ 0\end{bmatrix} = \begin{bmatrix}60 \\ 60\end{bmatrix}`.

After the collision, mass of the combined object becomes
`m = 20 + 30 = 50`. Divide momentum by the mass of the combined object to find the velocity of the combined object:

`\begin{aligned}v &= (p)/(m) = \frac{\begin{bmatrix}60 \\ 60\end{bmatrix}}{50} = \begin{bmatrix}1.2 \\ 1.2\end{bmatrix}\end{aligned}`.

Speed is the magnitude of velocity. The speed of this object would be:

`\begin{aligned}v &= \sqrt{1.2^(2) + 1.2^(2)} \approx 1.7\; {\rm m\cdot s^(-1)}\end{aligned}`.

The angle between this velocity and the first component can be found as the arctangent of the ratio between second and the first component:

`\displaystyle \arctan \left((1.2)/(1.2)\right) = 45^(\circ)`.

In other words, this velocity is
`45^(\circ)` east from north.