An explosion breaks an object into two pieces, one of which has 2.30 times the m
ID: 1347355 • Letter: A
Question
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)
Explanation / Answer
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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