3. As shown in the figure below, object m 1 = 1.45 kg starts at an initial heigh

Question

3. As shown in the figure below, object m1 = 1.45 kg starts at an initial height h1i = 0.340 m and speed v1i = 4.00 m/s, swings downward and strikes (in an elastic collision) object m2 = 4.55 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.)

____________m/s (m1)

___________ m/s (m2)

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

___________m (m1)

___________m (m2)

Explanation / Answer

here,

m1 = 1.45 kg

h1 = 0.34 m

v1i = 4 m/s

m2 = 4.55 m/s

a)

the velocity of m1 before collison be u1

using conservation of energy

0.5 * m1*v1i^2 + m1*g*h = 0.5 * m1 * u1^2

0.5*1.45*4^2 + 1.45*0.34*9.8 = 0.5 * 1.45*u1^2

u1 = 4.76 m/s

the speed of m1 before collison is 4.76 m/s

b)

let the final speeds be v1 and v2

using conservation of momentum

m1*u1 = m1*v1 + m2*v2

1.45 * 4.76 = 1.45*v1 + 4.55 * v2 ...(1)

using conservation of energy

0.5*m1*u1^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2

1.45*4.76^2 = 1.45*v1^2 + 4.55*v2^2 ...(2)

from (1) and (2)

v1 = - 2.5 m/s

v2 = 2.3 m/s

the final velocity are - 2.5 m/s and 2.3 m/s

c)

let the height for m1 be h1

using third equation of motion

v1^2 - u1^2 = 2 * g * h1

2.5^2 = 2 * 9.8 * h1

h1 = 0.32 m

let the height for m2 be h2

using third equation of motion

v2^2 - u2^2 = 2 * g * h2

2.3^2 = 2 * 9.8 * h2

h2 = 0.27 m

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