Question

A 15-gram stone is shot horizontally at a 3.0 kg wooden block hooked to a relax 250 N/m spring (the other end of the spring is fixed against a wall, the spring at the block sitting on a horizontal frictionless table). It is observed that the stone bounces back with a speed of 10 m/s, and that the blcok compresses the spring by 4.0 cm.

1a. What was the speed of the stone just before collision?

1b. Calculate the energy lost in the collision and express it as a percentage of the initial energy.

A 3.0 kg chunk of ice sliding at 12 m/s catches up with another chunk of ice, a 2.0 kg mass, already moving at 2.0 m/s on a frictionless surface. The chunks stick together and move up a frictionless hill. At the top of the hill, that is 2.0 m high, a spring of 1000 N/m is waiting to be compressed. Calculate the mazimal compression of the spring

Answer 1

compresses 4 cm

so get speed after collision

0.5*3*v^2= 0.5*250*0.04^2

v=0.3651 m/s

now apply conservation of momentum

0.015*v= -0.015*10 +3*0.3651

1 a.v=63.03 m/s

1b. energy before = o.5*0.015*63.03^2=29.7955 J

Energy after = 0.5*o.o15*10^2 +0.5*250*0.04^2 = 0.95 J

Loss = 29.7955-0.95 = 28.8455 J which is 96.8 %

Second question 3*12+2*2 = 5*v

v=8 m/s

now using energy conservation

o.5*5*8^2= 5*9.8*2+ KE on the top

KE on the top is =62 J Which is use to compress the spring

62= 0.5*1000*x^2

x=0.3521 m