please complete this with steps. A ball of mass 0.540 kg moving in the +X direct

Question

please complete this with steps.

A ball of mass 0.540 kg moving in the +X direction with a speed of 3.80 m/s and collides with another ball of mass 0.340 kg. The head-on collision is perfectly elastic. What is the velocity of these balls after the collision when the
a) second ball is at rest before the collision?
b) second ball is moving in the +X direction with a speed of 1.80 m/s before the collision? c) second ball is moving in the -X direction with a speed of 1.80 m/s before the collision?

2. Let f(x,y) = (1-x4-v4) 1/4 and let S be the surface given by the graph z-/(x,y) )1/4 and let S be the surface given by the graph z f(x,y) a) (1 pt) What is the domain of /? What happens to the values of f as the point (x,y) b) (I pt) Make a sketch of the surface S in 3-dimensional space, and also a (separate) c) (1 pt) Let C2 denote the curve in xy-plane given by 2-3/4 cos (t): sin (2t)·Let C gets near the edge of this domain? sketch of the contour plot of f denote the curve on the surface S which whose projection on to the xy-plane is C2 Find the parametric equations r = r(t) for C. d) (1 pt) Add to the sketches you gave in in part (b) i) C in the graph of S and ii) C2 on the contour plot of f e) (2 pts) Let z(t) denote the z-component of parametric equations r = r(t) of C you found in part (c). Find the points where z(t) has its local maxima and minima, and add these in to the sketch in part (b). 1) (1 pt) Set up the function h(t) which gives the square of the distance from the origin to a variable point on the curve C2 and then find the local maxima and minima of h(t).

Explanation / Answer

by conservation of momentum

initial momentum = final momentum

when second ball is in rest

0.54 * 3.8 + 0 = 0.54 * v1 + 0.34 * v2 ----- (1)

by conservation of energy

initial energy = final energy

0.5 * 0.54 * 3.8^2 + 0 = 0.5 * 0.54 * v1^2 + 0.5 * 0.34 * v2^2 ------- (2)

on solving equation 1 and 2 we'll get

v1 = 0.863 m/s

v2 = 4.66 m/s

a) speed of 0.54 kg mass = 0.863 m/s

a) speed of 0.34 kg mass = 4.66 m/s

when second ball is also moving with +1.8 m/s

0.54 * 3.8 + 0.34 * 1.8 = 0.54 * v1 + 0.34 * v2 ----- (3)

0.5 * 0.54 * 3.8^2 + 0.5 * 0.34 * 1.8^2 = 0.5 * 0.54 * v1^2 + 0.5 * 0.34 * v2^2 ----- (4)

on solving 3 and 4 we'll get

v1 = 2.25 m/s

v2 = 4.25 m/s

b) speed of 0.54 kg mass = 2.25 m/s

b) speed of 0.34 kg mass = 4.25 m/s

when second ball is also moving with -1.8 m/s

0.54 * 3.8 - 0.34 * 1.8 = 0.54 * v1 + 0.34 * v2 ----- (5)

0.5 * 0.54 * 3.8^2 + 0.5 * 0.34 * 1.8^2 = 0.5 * 0.54 * v1^2 + 0.5 * 0.34 * v2^2 ----- (6)

on solving 5 and 6 we'll get

v1 = -0.527 m/s

v2 = 5.07 m/s

c) speed of 0.54 kg mass = -0.527 m/s

c) speed of 0.34 kg mass = 5.07 m/s

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