Question

Please help ! Grades are due tmr and I just need a few questions !

A 17 kg ball of clay travelling at 14 m/s collides with a 7 kg wooden block sitting on a frictionless
surface. After the collision the clay and the block stick together. How fast are they moving after the
collision?

Answer 1

9.92 m/s

Step-by-step explanation:

We can use conservation of momentum to solve this problem. The total momentum of the system before the collision is:

p1 = m1 * v1 + m2 * v2,

where m1 and v1 are the mass and velocity of the clay ball before the collision, and m2 and v2 are the mass and velocity of the wooden block before the collision. Substituting the given values, we get:

p1 = 17 kg * 14 m/s + 7 kg * 0 m/s = 238 kg m/s.

After the collision, the clay and the block stick together and move as a single object with a combined mass of:

m = m1 + m2 = 17 kg + 7 kg = 24 kg.

Let the final velocity of the combined object be vf. Then, the conservation of momentum principle tells us that the total momentum of the system after the collision is:

p2 = m * vf.

Since momentum is conserved, we have p1 = p2, which gives us:

238 kg m/s = 24 kg * vf.

Solving for vf, we get:

vf = 238 kg m/s / 24 kg = 9.92 m/s.

Therefore, the clay and the block move together at a speed of 9.92 m/s after the collision.