Momentum and Energy 3. A new event has been proposed for the winter Olympics in

Question

Momentum and Energy 3. A new event has been proposed for the winter Olympics in Russia. As seen in the picture, an athlete will sprint and then leap unto a 20kg bobsled. The person and bobsled will travel down a 50m long ice covered ramp sloped at 20°, and into a carefully calibrated spring of k-2000 N/m. The athlete who compresses the spring by the most wins the gold! Electra, who's mass is 40kg has been friction and air resistance for this problem. 20° training and can hit the sled at a speed of 12 m/s. Ignore NOTE: When Electra collides with the sled, some system energy will be lost to Etherma which we are NOT keeping track of below. Each value in the table includes Electra AND the sled AND the spring. PEgravity PEsering Total System E Time KE Before sled collision (Electra+Sled) Just after sled collision (Electra+ Sled) At bottom of hill before spring contact-l At spring maximum compression

Explanation / Answer

Mass of sled = m = ­­20kg

Mass of Electra = M = 40kg

d=50m

k=2000N/m

vi=12m/s

=20deg

Before sled collision:

KE=1/2mvi^2 = ½*40*12^2 = 2880 J

PEgravity = (m+M)gh=(m+M)gdsin =(20+40)*9.8*50*sin20 = 10055.4 J

PEspring =1/2kx^2 =1/2*2000*0^2=0 J

Total system E = KE + PEgravity + PEspring = 2880 + 10055.4 + 0 = 12935.4 J

Just after sled collision:

By law of conservation of linear momentun

Pf = pi

Mvi=(m+M)*vf1

40*12 = (20+40)vf

Vf1= 8.0 m/s

KE=1/2(m+M)vf^2 = ½*(20+40)*8^2 = 1920 J

PEgravity = (m+M)gh=(m+M)gdsin =(20+40)*9.8*50*sin20 = 10055.4 J

PEspring =1/2kx^2 =1/2*2000*0^2=0 J

Total system E = KE + PEgravity + PEspring = 1920 + 10055.4 + 0 = 11975.4 J

At the bottom of hill before spring contract:

vf2^2=vf1^2-28h

vf2^2=vf1^2-28dsin20

vf2^2=8.0^2 -2*9.8*-50sin20

vf2=19.98m/s

KE=1/2(m+M)vf2^2 = ½*(20+40)*19.98^2 = 11976.01J

PEgravity = mgh=40*9.8*0 = 0.0 J

PEspring =1/2kx^2 =1/2*2000*0^2=0 J

Total system E = KE + PEgravity + PEspring = 11976.01 + 0.0 = 11976.01 J

At spring maximum compression :

Use law of conservation of energy

KEi= spring PEf

11976.01 =1/2*2000*x^2

x=3.46m

after spring compression velocty of sled +Electra = vf3 = 0 m/s

KE=1/2(m+M)vf3^2 = ½*(20+40)*0.0^2 = 0 J

PEgravity = mgh=40*9.8*0 = 0.0 J

PEspring =1/2kx^2 =1/2*2000*3.46^2=0 J

Total system E = KE + PEgravity + PEspring = 11976.01 + 0.0 = 11971.6 J

Related Questions

Navigate

• Browse (All)
• Browse M
• Subjects

---

---

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