A spring with a spring constant of 4.0 x 10^4 N/m is compressed .0633 meters and

Question

A spring with a spring constant of 4.0 x 10^4 N/m is compressed .0633 meters and a 0.1kg ball is placed on it. The axis of the spring is along a line 30 degrees above the horizontal.

1) Write the energy conservation equations for the ball at the instants when a) the spring is released, b) the ball leaves the spring, and c) when the ball is at the maximum height. Ignore the small height difference of the compressed spring.

2) Calculate the velocity at which the ball leaves the spring, the maximum height reached and the time to reach the maximum height.

A 0.2kg block is fired strait upward so that it is moving at 2 m/s upward when it collides and sticks to the first ball 3.0 seconds after it left the spring.

3) Calculate the velocity components of the ball just before the collision (at 3 seconds after it left hte spring)

4) Set up the momentum equations for the collision of the ball and block.

5) Solve for the components of the velocity of the combined ball and block just after collision.

Explanation / Answer

1) 0.5*40000*0.0633^2 = 0.5*m*u*u = 80.1378 J

speed of ball u = 40.03 m/s

at max height energy of ball = mghmax + 0.5*m*(ucos30)^2 = 80.1378 J

2)

u= 40.03 m/s

Hmax = (usin30)^2/2g = 20.04 m

t = usin30 / g = 2.042 sec

3)
for ball 1 Vx = ucos 30 = 34.66 ms
Vy = usin30 - 3g = -9.385 m/s( moving downwards)

4) 5)

horizontal direction

0.1*34.66 + 0 = (0.1+0.2)*Vcx
Vcx = 11.5 m/s

vertical direction

0.1*(-9.385) + 0.3*2 = (0.1+0.2)*Vcy

Vcy = -1.12 m/s(moving downward)

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.