In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

KE1 = KE2

The kinetic energy of the system before the collision is solved below.

KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²

This value should also be equal to KE2, which can be calculated using the conditions after the collision.

KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)

The value of x from the equation is 17.16 cm/s.

Hence, the answer is 17.16 cm/s.

6 0

3 years ago

Problem: A barbell consists of two small balls, each with mass m at the ends of a very low mass rod of length d. The barbell is

zepelin [54]

Answer:

mass of ball 1=m1

mass of ball 2=m2

velocity of ball=r1w1

velocity of ball 2=r2w2

Total angular momentum=m1*v1+m2*v2

but

v1=r1*w1

v2=r2*w2

Substitute values in above equation

Total angular momentum of the system=m1*r1*w1+m2*r2*w2

7 0

2 years ago

Small rockets are used to make small adjustments in the speed of satellites. One such rocket has a thrust of 42 N. If it is fire

Over [174]

To solve this problem we will apply the concepts related to Newton's second law that relates force as the product between acceleration and mass. From there, we will get the acceleration. Finally, through the cinematic equations of motion we will find the time required by the object.

If the Force (F) is 42N on an object of mass (m) of 83000kg we have that the acceleration would be by Newton's second law.

$F = ma \rightarrow a = \frac{F}{m}$