Question

In: Physics

An explosion breaks an object into two pieces, one of which has 2.30 times the mass...

An explosion breaks an object into two pieces, one of which has 2.30 times the mass of the other. If 7700 J were released in the explosion, how much kinetic energy did each piece acquire(a) Calculate the velocity of the target ball after the collision.

heavier piece? ___J

lighter piece? ____J

A softball of mass 0.220 kg that is moving with a speed of 8.0 m/s (in the positive direction) collides head-on and elastically with another ball initially at rest. Afterward it is found that the incoming ball has bounced backward with a speed of 6.3 m/s.

(a) Calculate the velocity of the target ball after the collision
____m/s
(b) Calculate the mass of the target ball.
____ kg

A 707 kg car stopped at an intersection is rear-ended by a 1830 kg truck moving with a speed of 11.5 m/s. If the car was in neutral and its brakes were off, so that the collision is approximately elastic, find the final speed of both vehicles after the collision.
____ m/s (car)
____ m/s (truck)

Solutions

Expert Solution

KE1 = P1^2/2m1


KE2 = p2^2/2m2

KE1 + KE2 = 7700


P1^2/2m1 + p2^2/2m2


P1 = P2 = P


m1 = 2.3 m2

P^2/2*2.3 m2 + P^2/m2 = 7700

(P^2/2m2) *(1/2.3 + 1) = 7700

P^2/2m2 = 5366.67 J

KE of second = P2^2/2m2 = 5366.67 J


KE of the first piece = KE1 = 7700-KE2 = 2333.33 J


++++++++++

m1 = 0.22 kg                m2 = ? kg

u1 = 8                   u2 = 0

after collision


v1 = -6.3              v2 = ?

from momewntum conservation

m1*u1 + m2*u2 = m1*v1 + m2*v2

0.22*8 + m2*0 = -0.22*6.3 + m2*v2 .............(1)

from energy conservation


0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2


0.22*8^2 = 0.22*6.3^2 + m2*v2^2............(2)


solving 1 & 2


m2 = 1.85 kg

v2 = 1.7 m/s


++++++++++


m1 = 707      u1 = 0

m2 = 1830      u2 = 11.5 m/s

after collision


v1 = ?          v2 = ?

from momewntum conservation

m1*u1 + m2*u2 = m1*v1 + m2*v2

1830*11.5 = 707*v1 + 1830*v2 .............(1)

from energy conservation


0.5*m1*u1^2 + 0.5*m2*u2^2 = 0.5*m1*v1^2 + 0.5*m2*v2^2


1830*11.5^2 = 707*v1^2 + 1830*v2^2............(2)


solving 1 & 2

v1 (car) = 16.5 m/s

v2 (truck) = 5.1 m/s


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