A small block of mass m 1 = 0.420 kg is released from rest at the top of acurved

Question

A small block of mass m1 = 0.420 kg is released from rest at the top of acurved-shaped, frictionless wedge of mass m2 =3.00 kg, which sits on a frictionless horizontal surface. When theblock leaves the wedge, its velocity is measured to bev1 = 3.20 m/s tothe right. (a) What is the velocity of the wedge after the block reachesthe horizontal surface?
1 m/s to the left

(b) What is the height h of the wedge?
2 m A small block of mass m1 = 0.420 kg is released from rest at the top of acurved-shaped, frictionless wedge of mass m2 =3.00 kg, which sits on a frictionless horizontal surface. When theblock leaves the wedge, its velocity is measured to bev1 = 3.20 m/s tothe right. (a) What is the velocity of the wedge after the block reachesthe horizontal surface?
1 m/s to the left

(b) What is the height h of the wedge?
2 m

Explanation / Answer

mass of the block, m1 = 0.42 kg mass of the wedge, m2 = 3 kg velocity of block , v1 = 3.2 m/s (a) momentum of the wedge = momentum of the block m2 v2 = m1v1 3 * v2 = 0.42 * 3.2 velocity of the wedge,v2 = 0.448 m/s ( left ) (b) Potential energy of the block at the highest point = kineticenergy of the block at the lowes point m1 g h = ( 1/2 ) m1(v1)2 h = ( v1 )2 / 2 g = ( 3.2 )2 / ( 2 * 9.8) = 0.52m h = ( v1 )2 / 2 g = ( 3.2 )2 / ( 2 * 9.8) = 0.52m

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