Question

A ball of unknown mass m is tossed straight up with initial speed v. At the moment it is released, the ball is a height h above a spring-mounted platform, as shown in the figure below. The ball rises, peaks, and falls back toward the platform, ultimately compressing the spring a maximum distance d from its relaxed position. Assume that the spring is perfectly ideal with spring constant k, and that the mass of the spring and platform is negligible.

What is the mass m, assuming that there is no friction or air resistance? Using g to represent the acceleration due to gravity, enter an expression for m in terms of g, h, d, k, and v.

Using g to represent the acceleration due to gravity, enter the expression for m in terms of g, h, d, k, and v.

Answer 1

The height reached by a body thrown stright up with initial velocity u = u²/2g

Here the initial velecity is given as v so the hieght reached is v²/2g

Total height reached above the spring is h + v²/2g as the body is thrown from a height h above the spring. At this height the velocity of the body is 0.

Now the body starts downward descent. The velocity of impact v₀is

v² = u² + 2as ------(u = 0, a = g, s = h + v²/2g)

v_{0 =}

v_{0 =}

At impact, two forces act on the body. The weight mg in downward direction, and the restoring force kx of the spring in the upward direction, where x is the displacement. The net force is the differencce of the 2 forces. Still the body continues to move in the downward direction compressing the spring.

net force = m a where a is the acceleration of the body after impact of the spring.

mg-kx = ma

a = (mg-kx)/m

Assume that the body moves from the point of impact until the point of full compression in small steps of . The corresponding acceleration at each point is

a₁= (mg-kx)/m

a₂= (mg-k2x)/m

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a_n= (mg-knx)/m

a_n = ng - (after approximating 1+2+3+...+n = n²/2 as n is a very large number)

Use equation v² = u² + 2aS and finding corresponding velocities at each point we get

v₁² = v₀² + 2 a₁x

v₂² = v₁² + 2 a₂x = v₀² + 2 a₁x + 2 a₂x = v₀² + 2 x (a₁ + a₂)

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.

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v_n² = v₀² + 2 x (a₁ + a_{2 + ...+} a_n )

But v_n = 0 as the spring is fully compressed, so

0 = v₀² + 2 x

nx = d since we divided d into small steps, so

0 = v₀² + 2dg - kd²/m

kd²/m = v₀² + 2dg

m = kd²/(v₀² + 2dg)

replacing value of v₀ we get

m = kd²/( 2gh + v²+ 2dg)