1.a ball rolls horizotally with a speed of 7.8 m/s off the edge of tall platform

ID: 1426707 • Letter: 1

Question

1.a ball rolls horizotally with a speed of 7.8 m/s off the edge of tall platform.

part A if the ball and 7.5 m from the point on the ground directly beow the edge of the platform, what is the height of the palform?h=__m 2.. a

2. 1100 kg automobile travels at 80 km/h.

part A. whats the kinetic energy?___J

part b what network would be required to bring it to stop?

3. A 4.2 kg ball with a velocity of 3.0 m/s in the +x-direction collides head-on elastically with a stationary 1.9 kg ball.

Part A What is the speed of the first ball after the collision? v = m m/s

Part B What is the direction of the velocity of the first ball after the collision? +x-direction or x-direction

Part C What is the speed of the second ball after the collision?

Part C What is the direction of the velocity of the second ball after the collision?n+x-direction or x-direction

1 ) the time taken by the ball to reach ground

T = distance / speed =7.5 / 7.8

= 0.96 s

Height ( vertical distance traveled ) = ut + gt^2/2

Since initial vertical velocity u = 0

h = gt^2 /2 = 9.81 * 0.96^2 /2

= 4.52 m

2) a ) kinetic energy E k = m v^2 /2

Given m = 1100 kg

v = 80 km /h = 22.2 m/s

E k = 1100 × 22 2 ^2 /2

= 27.16 *10^4 J

2) b ) questions is not incomplete and wrong.

3 ) A) given

m1 = 4.2 kg u 1 = 3m/s

m 2 = 1.9 kg u 2 =0

For a elastic collision

final velocity v 1 = u 1 ( m1 - m2 )/(m1 + m 2) + u 2 (2* m2 /m1 + m2 )

v 1 = 3 ( 4.2 - 1.9 )/ (4.2 + 1.9 ) + 0

= 1.131 m/s

B) since m 1 > m 2

Direction will be + x

C) v2 = u 2 (m 2 - m1)/(m1 + m2) + u1(2m1/m1+m2)

= 0 + 3 ( 2*4.2 / 4.2+1.9)

4.13 m /s

D) direction is + x

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