1. A block of mass m = 3.5 {\ m kg}, moving on a frictionless surface with a spe
ID: 1394478 • Letter: 1
Question
1.
A block of mass m = 3.5 { m kg}, moving on a frictionless surface with a speed {{v}}_{{i}} = 6.1 { m m/s}, makes a perfectly elastic collision with a block of mass M at rest. After the collision, the 3.5 kg block recoils with a speed of {{v}}_{{f}} = 1.9 { m m/s}. In the figure, the blocks are in contact for 0.20 s. The average force on the 3.5 { m kg} block, while the two blocks are in contact, is closest to:
0 N
2. A block of mass m = 34 kg has a speed V and is behind a block of mass M = 79 kg that has a speed of 0.5 m/s . The surface is frictionless. The blocks collide and couple. After the collision, the blocks have a common speed of 0.9 m/s . In the figure, the loss of kinetic energy of the blocks due to the collision is closest to:
107 N 123 N 140 N 74 N0 N
2. A block of mass m = 34 kg has a speed V and is behind a block of mass M = 79 kg that has a speed of 0.5 m/s . The surface is frictionless. The blocks collide and couple. After the collision, the blocks have a common speed of 0.9 m/s . In the figure, the loss of kinetic energy of the blocks due to the collision is closest to:
21 J 93 J 2.0 J 17 J 36 JExplanation / Answer
1) From the Formula of impulse J = Favg*dt = m*(vf-(-vi)) = m*(vf+vi)
given that m = 3.5 kg
vf = 1.9 m/s
vi = 6.1 m/s
time of contact dt = 0.2 S
average force Favg = ?
Favg*dt = m*(vf+vi)
Favg*dt = 3.5*(1.9+6.1)
Favg *dt = 28
Favg = 28/dt = 28/0.2 = 140 N
2) From the law of conservation of momentum
momentum before collision = momentum after collision
(m*V)+(M*0.5) = (m+M)*0.9
(34*V)+(79*0.5) = (34+79)*0.9 = 101.7
34*V = 101.7-(79*0.5) = 62.2
V = 62.2/34 = 1.8 m/s
then loss in Kinetic energy of the blocks is KE_final - KE_inital
KE_initial = (0.5*m*V^2)+(0.5*M*0.5^2) = (0.5*34*1.8^2)+(0.5*79*0.5^2)
KE_initial = 64.955 J
KE_final = 0.5*(m+M)*0.9^2 = 0.5*(34+79)*0.9^2 = 45.765 J
loss of KE = KE_final - KE_initial = 45.765-64.955 =-19.19 J
which is closest to 21 J.....if 21J is not correct then try with 17J
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