As shown in the figure below, object m1 = 1.45 kg starts at an initial height h1

Question

As shown in the figure below, object m1 = 1.45 kg starts at an initial height h1i = 0.295 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.60 kg which is initially at rest. Determine the following.

(a) speed of m1 just before the collision. (m/s)

(b) velocity (magnitude and direction) of each ball just after the collision (Assume the positive direction is toward the right. Indicate the direction with the sign of your answer.)

m1 ___ (m/s)

m2 ___ (m/s)

(c) height to which each ball swings after the collision (ignoring air resistance)

m1 ___ m

m2 ___m

Explanation / Answer

m1 = 1.45 kg, h1i =0.295 m , v1i = 4m/s

m2 = 4.6 kg

(a) from kinematic equaitons

v^2 -u^2 = 2as

u1^2 - 4^2 = 2*9.8*0.295

u1 = 4.667 m/s

(b) from conservation of momentum

m1u1 +m2u2 = m1v1 +m2v2

1.45*4.667 +0 = 1.45*v1+4.6*v2 ..(1)

for elastic collision

v2 - v1 = u1 - u2

v2 - v1 = 4.667 ... (2)

from (1) and (2) we get

v1 = -2.43 m/s

v2 = 2.237 m/s

m1 moving with 2.43 m/s towards left

m2 moving with 2.237 m/s towards right

(c) h1 = v1^2/2g

h1 = 2.43^2/(2*9.8)

h1 = 0.3 m

h2 = 2.237^2/(2*9.8)

h2 = 0.255 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

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