Question

An automobile having a mass of 1,000 kg is driven into a brick wll in a safety test. The bumper behaves like a spring with constant 4.35 x 10 N/m and is compressed 2.45 cm as the car is brought to rest. What was the speed of the car before impact, assuming no energy is lost in the collision with the wall? m/s Need Help? Liina LmLǐ 8. O-41 points SerCP10 13.P013 Mr Notes Ask Your Teacher . A 10.0-g bullet is fired into, and embeds itself in, a 1.80-kg block attached to a spring compressed if the bulilet has a speed of 300 m/s just before it strikes the block this problem because of the inelastic collision between the bullet and block with a force constant of 21.3 N/m and whose mass is negligible. How far is the spring and the block slides on a frictionless surface? Note: You must use conservation of momentum in Need Help? LRosd N .-44 points SerCP10 13P017 My Notes Ask Your A 0.39-kg object connected to a light spring with a force constant of 22.2 N/m osciliates on a frictionless rest. horizontal surface. The spring is compressed 4.0 cm and released from (a) Determine the maximum speed of the object m/s (b) Determine the speed of the object when the spring is compressed 1.5 cm mVs (c) Determine the speed of the object as it passes the point 1.5 cm from the equililbrium position m/s (d) For what value of x does the speed equal one-half the maximum speed?

Please help with any or all problems I can’t seem to figure them out thanks

Answer 1

(7) Suppose the speed of the car before the collision = v m/s

So, kinetic energy of the car before collision, KE = (1/2)*m*v^2

Compression in the spring, x = 2.45 cm = 0.0245 m

Spring constant of the spring, k = 4.35 x 10^6 N/m

So, spring energy stored after the collision, SE = (1/2)*k*x^2

By conservation of energy –

KE = SE

=> (1/2)*m*v^2 = (1/2)*k*x^2

=> v^2 = [k*x^2] / m = [4.35 x 10^6 * 0.0245^2] / 1000

=> v = 1.616 m/s (Answer)

(8) Here, mass of the bullet = m1 = 10.0 g = 0.01 kg

Mass of the block, m2 = 1.80 kg

suppose the speed of bullet and the block after firing = V

Use conservation of momentum –

m1*v1 + m2*v2 = (m1+m2)*V

=> 0.01*300 + 1.80*0 = (0.01+1.80)*V

=> V = (0.01*300) / 1.81 = 1.66 m/s

Now, suppose the spring compresses a distance x meter after striking.

Apply conservation of energy –

(1/2)*(m1+m2)*V^2 = (1/2)*k*x^2

=> (m1+m2)*V^2 = k*x^2

=> 1.81*1.66^2 = 21.3*x^2

=> x^2 = (1.81*1.66^2) / 21.3

=> x = 0.484 m (Answer)