Question

A projectile launcher as shown in the Figure is used to launch 0.5 kg ball into the air. The launcher is connected to a 1kg mass and starts from rest with the spring compressed to a length of 15 cm. The un-stretched length of the spring is 75 cm. When the launcher is released from rest, the 1kg mass will impact the ball and the ball wil project in the air. Determine a suitable spring constant and angle 8 for ball to reach a maximum height of 20 m (measure from the current position of ball). The Coefficient of restitution between 1kg mass and ball is 0.8. Note: first determine an arbitrary angle and find the spring constant for that) 10.5kg) 75 cm 1kg 15 cm 0

Answer 1

We have the given data,

M=1kg, m=0.5kg, h=20m, e=0.8.

We have to find the spring constant and also an arbitrary angle which we will assume as 35°.

Here the system we consider is the spring, masses, and the earth.

Work = change in kinetic energy+ spring energy + gravitational energy = 0

We first solve for the final velocity of the 1kg mass named A, the system here is again consisting of spring, mass, and gravity. From law of conservation of momentum for a perfect elastic collision, we have the equation 1.

The equation below is the law of conservation of energy for a projectile.

W = Ak HAS + AU = 0 - 0

Now substituting all the values in the expansion we get the relation 2. From the law of conservation of momentum we have the equation 3 and for perfectly elastic collision we have the equation 4, substituting and solving both the equations we get the relation 7.

B=0.5kg. l = (75-15) • esino k 0 en we A + Mgh 2 A=1kg - Ikea Ocks. votes - K(75x102-15x10*2)2 +1.69.81). 10 d sin(359) k (3600,-675.21.5 Via "A X10-4) X10-2) BB 1t + 1* VA 5 MAVA + my MAYA +moves VA + "' - Vzt ve' for perfectly elastic collision. @ /* VA' +0.5 Vé' VA = VA'+0.5 +0.5 V6' VA TV = Vot ve VA = Va'-A' from 6 50 VA' + 0.5 VB = VE-VA VA'= 0.25 Vis' -

But the coefficient of restitution is 0.8 and substituting the respective velocities we get the relation between the final velocity of the 1kg mass and velocity of the initial velocity if the ball after collision, now buy substituting the values in the conservation of energy equation for the ball with the relation 2, 8 in 1. We get k=123.94N/m

But e=0.8= VB - VA VA - Yo e=0.8 = VB - 0.25 VA - VA 25 We'lor van VA)) V 0.8 VA 0.8VA = Va' (1-0-25) ve' - (1-0.25) VA. = 16 VA GR I kf2+m mgh. < 0.5 * (K*0.36 – 6-7521)* *(44BADA => 0 = 1 2 m B 22 1 k (75x10-2_15x10 2300 0 a 16/15 + 0.5*9.81 *20 Solving for k get. K - 123.94 N/m we