Alarge 10.kg medicine ball is caught by a 70.kg student on the track team. if the ball was moving at 4.0 m/s, how fast will the student be moving after catching the ball?

v2 = ?

m1 = 10kg

m2 = 70kg

v1 = 4m/s

E1 = E2

E1 = 1/2 * m1 * v1^2 = 1/2 * 10kg * 4m/s^2 = 80J

E2 = 1/2 * m2 * v2^2 = 80 J

v2 = √(E2/(2 * m2)) = √(80J/(2 * 70kg)) = about 0.76m/s

V=0.5m/s

Explanation:

A large 10.kg medicine ball is caught by a 70.kg student on the track team. If the ball was moving at 4.0 m/s, how fast will the student be moving after catching the ball?

taking it stepwisely

This problem can be solved with the knowledge of momentum.

the total momentum before collision must be equal to the total momentum after collusion.

This is an inelastic collision because the object and the boy will stick together after catching the boy.

Therefore

m1u1+m2u2= (m1+m2)V

m1=mass of the medicine ball 10kg

m2= mass of the student,70kg

U1=4m/s

u2=0m/s

V=common velocity after the student catches the medicine ball

10*4+70*0= (10+70)V

40=80V

V=40/80

V=0.5m/s

The student will move with a speed of 0.5m/s after catching the ball.

the results is a decrease in air temperature with an increase of altitude.

explanation:

need to see circuit to get time constant