Question

A box initially at rest is struck by a mass of clay moving at 20 m/s. The clay sticks to a 10 kg box (initially at rest) causing it to move at 5 m/s. Calculate the mass of clay that strikes the box

I NEED HELP ASAP!!!

Answer 1

E

Step-by-step explanation:

Answer 2

Approximately
`3.33\; \rm kg` if there's no friction on the box.

Step-by-step explanation:

This problem can be solved with the conservation of momentum.

The momentum
`p` of an object is equal its mass
`m` times its velocity
`v`. That is:
`p = m \cdot v`.

In a collision between two objects, the sum of their momentum will stay the same. As long as all external forces (e.g., gravity) on the two objects are balanced, momentum will be conserved even if the two objects eventually stuck to each other.

Let the mass of the clay be
`x\; \rm kg`.

Before the collision, the clay is moving at
`20\; \rm m \cdot s^(-1)` while the box isn't moving/

• Momentum of the clay:
  `(20\, x)\; \rm kg \cdot m \cdot s^(-1)`.
• Momentum of the box:
  `0\; \rm kg \cdot m \cdot s^(-1)` since the velocity of the box is zero.
• Sum of momentum before collision:
  `(20\, x)\; \rm kg \cdot m \cdot s^(-1)`.

After the collision, the clay is stuck to the box. As a result, both the box and the clay will be moving at
`5\; \rm m \cdot s^(-1)` (the direction of travel is the same as that of the clay before the collision.)

• Momentum of the clay:
  `(5\, x)\; \rm kg \cdot m \cdot s^(-1)`.
• Momentum of the box:
  `10\; \rm kg * 5\; \rm m \cdot s^(-1) = 50\; \rm kg \cdot m \cdot s^(-1)`.
• Sum of momentum after collision:
  `(5\, x + 50)\; \rm kg \cdot m \cdot s^(-1)`.

If there's no friction or other unbalanced external force on the clay or the box, momentum will be conserved in the collision. As a result,
`(20\, x)\; {\rm kg \cdot m \cdot s^(-1)} = (5\, x + 50)\; {\rm kg \cdot m \cdot s^(-1)}`.

Solve this equation for
`x`, the mass of the clay (in kilograms:)
`x \approx 3.3`.

In other words, the mass of the clay is approximately
`3.3\;\rm kg`.