can some one please help using impulse and momentum;the answer as professor indi

Question

can some one please help using impulse and momentum;the answer as professor indicated is 48.3 cm; please show detialed steps and pictures.

5. A block with mass m1 = 900 g is held on a horizontal frictionless track against a spring with k = 120 N/m, which is initially compressed 10.0 cm. The block is released and it collides with a second block with m2 250 g. After the collision, both blocks slide over a rough surface where the coefficient of kinetic friction between the each block and the surface is 0.300. If the collision is elastic, find the distance between the blocks when they are both at rest. m1 rough surface

Explanation / Answer

here,

m1 = 0.9 kg

spring constant , K = 120 N/m

x= 0.1 m

let the speed before the collison be u

using conservation of energy

0.5 * k * x^2 = 0.5 *m1 * u^2

120 * 0.1^2 = 0.9 * u^2

u = 1.15 m/s

m2 = 0.25 kg

using conservation of momentum

m1 * u = m1 * v1 + m2 * v2

0.9 * 1.15 = 0.9 * v1 + 0.25 * v2 ....(1)

and

using conservation of kinetic energy

0.5 * m1 * u^2 = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2

0.9 * 1.15^2 = 0.9 * v1^2 + 0.25 * v2^2 .....(2)

from (1) and (2)

v1 = 0.65 m/s

v2 = 1.8 m/s

accelration due to friction , a = - uk * g = - 2.91 m/s^2

the distance travelled by 1 , s1 = v1^2 /2g = 0.07 m = 7.18 cm

the distance travelled by 2 , s2 = v2^2 /2g

s2 = 1.8^2 /( 2 * 9.81) = 16.5 cm

the distance between them , s = s2 - s1 = 9.33 cm

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