or maximally inelastic)? Justify your answer l Is this (elastic, inelastic A pen

Question

or maximally inelastic)? Justify your answer l Is this (elastic, inelastic A pendulum bob of mass 4 kg is released from height H and swings down to collide with a stationary block of mass 6 kg. Right before the collision the bob is * 32. oving to the left at speed 5 m/s. Just after the collision, the block is traveling to the left at speed 3 m/s. The block then slides over a very long, rough surface with a coefficient of kinetic frict past the beginning of the rough section. tion ?k 0.2, bringing the block to a stop at a point Ax K 0.2 6kg4kg_ Ar a) Find the height H from which the bob was released. b) Find the velocity (magnitude and direction) of the bob right after the collision. o) What is the bob-block system's kinetic energy just before the colision. d) What is the bob-block system's kinetic energy just after the collision. e) Is this collision maximally inelastic, inelastic, or elastic? Explain f) How far will the block slide across the rough section of track before it comes to a halt? lidor down a frictionless track so that its center

Explanation / Answer

here,

32)

mass of bob , m1 = 4 kg

mass of block , m2 = 6 kg

initial speed of bob , u1 = - 5 m/s

final speed of block , v2 = - 3 m/s

a)

the height , H = u1^2 /2g

H = 5^2 /(2 * 9.81) m

H = 1.28 m

b)

let the velocity of bob after the collison be v1

using conservation of momentum

m1 * u1 = m1 * v1 + m2 * v2

4 * (-5) = 4 * v1 + 6 * (-3)

solving for v1

v1 = - 0.5 m/s

the final velocity of bob is 0.5 m/s to the left

c)

the bob-block kinetic energy before the collison , KEi = 0.5 * m1 * u1^2

KEi = 0.5 * 4 * 5^2 J = 50 J

d)

the bob-block kinetic energy after the collison , KEf = 0.5 * m1 * v1^2 + 0.5 * m2 * v2^2 +  

KEf = 0.5 * 4 * 0.5^2 + 0.5 * 6 * 3^2 J = 27.5 J

e)

as KEi is not equal to KEf

the collison is elastic

f)

uk = 0.2

accelration , a = - uk * g = - 1.96 m/s^2

the distance travelled before stopping , s = v2^2 /( 2a)

s = 3^2 /( 2 * 1.96) m = 2.3 m

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